How the answer of this for loop is 5?
I was practicing my C skills online. And I got a question like:
What is the output of this program?
void main(){
int i;
for(i=1;i++<=1;i++)
i++;
printf("%d",i);
}
The answer was 5. But I thought the for loop would execute endlessly. As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
c
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
|
show 2 more comments
I was practicing my C skills online. And I got a question like:
What is the output of this program?
void main(){
int i;
for(i=1;i++<=1;i++)
i++;
printf("%d",i);
}
The answer was 5. But I thought the for loop would execute endlessly. As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
c
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
The proper declarations formainareint main (void)andint main (int argc, char **argv)(which you will see written with the equivalentchar *argv). note:mainis a function oftype intand it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?
– David C. Rankin
Nov 22 '18 at 4:51
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29
|
show 2 more comments
I was practicing my C skills online. And I got a question like:
What is the output of this program?
void main(){
int i;
for(i=1;i++<=1;i++)
i++;
printf("%d",i);
}
The answer was 5. But I thought the for loop would execute endlessly. As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
c
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
I was practicing my C skills online. And I got a question like:
What is the output of this program?
void main(){
int i;
for(i=1;i++<=1;i++)
i++;
printf("%d",i);
}
The answer was 5. But I thought the for loop would execute endlessly. As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
c
c
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
asked Nov 22 '18 at 4:45
MrAlphaMrAlpha
4619
4619
Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
The proper declarations formainareint main (void)andint main (int argc, char **argv)(which you will see written with the equivalentchar *argv). note:mainis a function oftype intand it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?
– David C. Rankin
Nov 22 '18 at 4:51
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29
|
show 2 more comments
Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
The proper declarations formainareint main (void)andint main (int argc, char **argv)(which you will see written with the equivalentchar *argv). note:mainis a function oftype intand it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?
– David C. Rankin
Nov 22 '18 at 4:51
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29
Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
The proper declarations for
main are int main (void) and int main (int argc, char **argv) (which you will see written with the equivalent char *argv). note: main is a function of type int and it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?– David C. Rankin
Nov 22 '18 at 4:51
The proper declarations for
main are int main (void) and int main (int argc, char **argv) (which you will see written with the equivalent char *argv). note: main is a function of type int and it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?– David C. Rankin
Nov 22 '18 at 4:51
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29
|
show 2 more comments
3 Answers
3
active
oldest
votes
This is explained by the sequence of events:
i = 1; // for init
i++ <= 1 ? true // for condition, i evaluates as 1 but is made into i = 2 after the expression
i++; // inside for body, makes i = 3
i++; // for increment, makes i = 4
i++ <= 1 ? false // for condition again, i evaluates as 4 but is made into i = 5 after the expression
// condition is false, for loop ends, i = 5
Perhaps you are forgetting that the for condition, although false, is still executed to verify that before the program decides the loop is terminated.
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
add a comment |
i is incremented four times before the for loop exits.
- Twice in the
forcondition check (formallycond_expression). Since you are using postfix increment, the first timeiwill be1when the condition is checked. It is only on the second check,iwill be2. - Once in the increment (formally
iteration_expression) - Once more in the body (formally
loop_statement) of theforloop.
As the initial value of i is 1, this brings its value to 5.
Note:
- cond_expression is evaluated before the loop body. If the result of the expression is zero, the loop statement is exited immediately.
- iteration_expression is evaluated after the loop body and its result is discarded. After evaluating iteration_expression, control is transferred to cond_expression.
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
add a comment |
As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
How can i never be less than or equal to 1 when it's initialised with a value of 1?
Did you intend to pre-increment i?
That would be:
for (i=1; ++1 <= 1; i++)
But you seem to misunderstand the nature of the conditional expression here, with a for loop you are repeating a block of code until a condition is met, in this case do code while i++ <= 1.
If you do
for (i=1; ++i <= 1; i++)
i will be > 1 on the first iteration and nothing will happen.
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
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votes
active
oldest
votes
This is explained by the sequence of events:
i = 1; // for init
i++ <= 1 ? true // for condition, i evaluates as 1 but is made into i = 2 after the expression
i++; // inside for body, makes i = 3
i++; // for increment, makes i = 4
i++ <= 1 ? false // for condition again, i evaluates as 4 but is made into i = 5 after the expression
// condition is false, for loop ends, i = 5
Perhaps you are forgetting that the for condition, although false, is still executed to verify that before the program decides the loop is terminated.
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
add a comment |
This is explained by the sequence of events:
i = 1; // for init
i++ <= 1 ? true // for condition, i evaluates as 1 but is made into i = 2 after the expression
i++; // inside for body, makes i = 3
i++; // for increment, makes i = 4
i++ <= 1 ? false // for condition again, i evaluates as 4 but is made into i = 5 after the expression
// condition is false, for loop ends, i = 5
Perhaps you are forgetting that the for condition, although false, is still executed to verify that before the program decides the loop is terminated.
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
add a comment |
This is explained by the sequence of events:
i = 1; // for init
i++ <= 1 ? true // for condition, i evaluates as 1 but is made into i = 2 after the expression
i++; // inside for body, makes i = 3
i++; // for increment, makes i = 4
i++ <= 1 ? false // for condition again, i evaluates as 4 but is made into i = 5 after the expression
// condition is false, for loop ends, i = 5
Perhaps you are forgetting that the for condition, although false, is still executed to verify that before the program decides the loop is terminated.
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
This is explained by the sequence of events:
i = 1; // for init
i++ <= 1 ? true // for condition, i evaluates as 1 but is made into i = 2 after the expression
i++; // inside for body, makes i = 3
i++; // for increment, makes i = 4
i++ <= 1 ? false // for condition again, i evaluates as 4 but is made into i = 5 after the expression
// condition is false, for loop ends, i = 5
Perhaps you are forgetting that the for condition, although false, is still executed to verify that before the program decides the loop is terminated.
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
answered Nov 22 '18 at 4:56
HavenardHavenard
17.9k22646
17.9k22646
add a comment |
add a comment |
i is incremented four times before the for loop exits.
- Twice in the
forcondition check (formallycond_expression). Since you are using postfix increment, the first timeiwill be1when the condition is checked. It is only on the second check,iwill be2. - Once in the increment (formally
iteration_expression) - Once more in the body (formally
loop_statement) of theforloop.
As the initial value of i is 1, this brings its value to 5.
Note:
- cond_expression is evaluated before the loop body. If the result of the expression is zero, the loop statement is exited immediately.
- iteration_expression is evaluated after the loop body and its result is discarded. After evaluating iteration_expression, control is transferred to cond_expression.
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
add a comment |
i is incremented four times before the for loop exits.
- Twice in the
forcondition check (formallycond_expression). Since you are using postfix increment, the first timeiwill be1when the condition is checked. It is only on the second check,iwill be2. - Once in the increment (formally
iteration_expression) - Once more in the body (formally
loop_statement) of theforloop.
As the initial value of i is 1, this brings its value to 5.
Note:
- cond_expression is evaluated before the loop body. If the result of the expression is zero, the loop statement is exited immediately.
- iteration_expression is evaluated after the loop body and its result is discarded. After evaluating iteration_expression, control is transferred to cond_expression.
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
add a comment |
i is incremented four times before the for loop exits.
- Twice in the
forcondition check (formallycond_expression). Since you are using postfix increment, the first timeiwill be1when the condition is checked. It is only on the second check,iwill be2. - Once in the increment (formally
iteration_expression) - Once more in the body (formally
loop_statement) of theforloop.
As the initial value of i is 1, this brings its value to 5.
Note:
- cond_expression is evaluated before the loop body. If the result of the expression is zero, the loop statement is exited immediately.
- iteration_expression is evaluated after the loop body and its result is discarded. After evaluating iteration_expression, control is transferred to cond_expression.
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
i is incremented four times before the for loop exits.
- Twice in the
forcondition check (formallycond_expression). Since you are using postfix increment, the first timeiwill be1when the condition is checked. It is only on the second check,iwill be2. - Once in the increment (formally
iteration_expression) - Once more in the body (formally
loop_statement) of theforloop.
As the initial value of i is 1, this brings its value to 5.
Note:
- cond_expression is evaluated before the loop body. If the result of the expression is zero, the loop statement is exited immediately.
- iteration_expression is evaluated after the loop body and its result is discarded. After evaluating iteration_expression, control is transferred to cond_expression.
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
edited Nov 22 '18 at 5:11
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
answered Nov 22 '18 at 4:49
P.WP.W
12.7k3944
12.7k3944
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
add a comment |
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
1
1
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
@MrAlpha: I update the answer. Please see if this clarifies.
– P.W
Nov 22 '18 at 5:04
add a comment |
As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
How can i never be less than or equal to 1 when it's initialised with a value of 1?
Did you intend to pre-increment i?
That would be:
for (i=1; ++1 <= 1; i++)
But you seem to misunderstand the nature of the conditional expression here, with a for loop you are repeating a block of code until a condition is met, in this case do code while i++ <= 1.
If you do
for (i=1; ++i <= 1; i++)
i will be > 1 on the first iteration and nothing will happen.
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
add a comment |
As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
How can i never be less than or equal to 1 when it's initialised with a value of 1?
Did you intend to pre-increment i?
That would be:
for (i=1; ++1 <= 1; i++)
But you seem to misunderstand the nature of the conditional expression here, with a for loop you are repeating a block of code until a condition is met, in this case do code while i++ <= 1.
If you do
for (i=1; ++i <= 1; i++)
i will be > 1 on the first iteration and nothing will happen.
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
add a comment |
As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
How can i never be less than or equal to 1 when it's initialised with a value of 1?
Did you intend to pre-increment i?
That would be:
for (i=1; ++1 <= 1; i++)
But you seem to misunderstand the nature of the conditional expression here, with a for loop you are repeating a block of code until a condition is met, in this case do code while i++ <= 1.
If you do
for (i=1; ++i <= 1; i++)
i will be > 1 on the first iteration and nothing will happen.
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
As the i will be incremented on each iteration and i will be never less than or equal to 1. how come 5 will be the output of this program?
How can i never be less than or equal to 1 when it's initialised with a value of 1?
Did you intend to pre-increment i?
That would be:
for (i=1; ++1 <= 1; i++)
But you seem to misunderstand the nature of the conditional expression here, with a for loop you are repeating a block of code until a condition is met, in this case do code while i++ <= 1.
If you do
for (i=1; ++i <= 1; i++)
i will be > 1 on the first iteration and nothing will happen.
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
answered Nov 22 '18 at 4:58
NunchyNunchy
815411
815411
add a comment |
add a comment |
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Possible duplicate of Undefined behavior and sequence points
– Ken White
Nov 22 '18 at 4:47
The proper declarations for
mainareint main (void)andint main (int argc, char **argv)(which you will see written with the equivalentchar *argv). note:mainis a function oftype intand it returns a value. See: C11 Standard §5.1.2.2.1 Program startup p1 (draft n1570). See also: See What should main() return in C and C++?– David C. Rankin
Nov 22 '18 at 4:51
@DavidC.Rankin What makes you think the OP is using a hosted system compiler? See What should main() return in C and C++?
– Lundin
Nov 22 '18 at 7:44
Why should it execute endlessly if you assume the condition will never be true? If your assumption was correct, it should not execute at all.
– Gerhardh
Nov 22 '18 at 8:24
@Lundin - I'm just a simple man, all I can do is read the standard unless the OP provides specifics. Perhaps we need to drop you same comment below each referencing What should main() return in C and C++??
– David C. Rankin
Nov 22 '18 at 18:29