Converting 4bytes to signed and unsigned ints
I want to convert a four-byte string to either int32 or uint32 in c++.
This answer helped me to write the following code:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
int v = *(unsigned int *) a.c_str();
int x = *(int32_t *) a.c_str();
int y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
However the outputs are all the same, where the int (or int32_t) values are correct, but the unsigned ones are wrong. Why does not the unsigned conversions work?
// output
int: -1442840406
uint: -1442840406
int32_t: -1442840406
uint32_t: -1442840406
Pythons struct.unpack, gives the right conversion
In [1]: import struct
In [2]: struct.unpack("<i", b"xaax00x00xaa")
Out[2]: (-1442840406,)
In [3]: struct.unpack("<I", b"xaax00x00xaa")
Out[3]: (2852126890,)
I would also like a similar solution to work for int16 and uint16, but first things first, since I guess an extension would be trivial if I manage to solve this problem.
c++ int
asked Nov 24 '18 at 19:38
user10668188user10668188
193
add a comment |
I want to convert a four-byte string to either int32 or uint32 in c++.
This answer helped me to write the following code:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
int v = *(unsigned int *) a.c_str();
int x = *(int32_t *) a.c_str();
int y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
However the outputs are all the same, where the int (or int32_t) values are correct, but the unsigned ones are wrong. Why does not the unsigned conversions work?
// output
int: -1442840406
uint: -1442840406
int32_t: -1442840406
uint32_t: -1442840406
Pythons struct.unpack, gives the right conversion
In [1]: import struct
In [2]: struct.unpack("<i", b"xaax00x00xaa")
Out[2]: (-1442840406,)
In [3]: struct.unpack("<I", b"xaax00x00xaa")
Out[3]: (2852126890,)
I would also like a similar solution to work for int16 and uint16, but first things first, since I guess an extension would be trivial if I manage to solve this problem.
c++ int
asked Nov 24 '18 at 19:38
user10668188user10668188
193
2
Because you're placing them all in (signed)intvariables.
– Rotem
Nov 24 '18 at 19:45
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49
add a comment |
I want to convert a four-byte string to either int32 or uint32 in c++.
This answer helped me to write the following code:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
int v = *(unsigned int *) a.c_str();
int x = *(int32_t *) a.c_str();
int y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
However the outputs are all the same, where the int (or int32_t) values are correct, but the unsigned ones are wrong. Why does not the unsigned conversions work?
// output
int: -1442840406
uint: -1442840406
int32_t: -1442840406
uint32_t: -1442840406
Pythons struct.unpack, gives the right conversion
In [1]: import struct
In [2]: struct.unpack("<i", b"xaax00x00xaa")
Out[2]: (-1442840406,)
In [3]: struct.unpack("<I", b"xaax00x00xaa")
Out[3]: (2852126890,)
I would also like a similar solution to work for int16 and uint16, but first things first, since I guess an extension would be trivial if I manage to solve this problem.
c++ int
asked Nov 24 '18 at 19:38
user10668188user10668188
193
I want to convert a four-byte string to either int32 or uint32 in c++.
This answer helped me to write the following code:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
int v = *(unsigned int *) a.c_str();
int x = *(int32_t *) a.c_str();
int y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
However the outputs are all the same, where the int (or int32_t) values are correct, but the unsigned ones are wrong. Why does not the unsigned conversions work?
// output
int: -1442840406
uint: -1442840406
int32_t: -1442840406
uint32_t: -1442840406
Pythons struct.unpack, gives the right conversion
In [1]: import struct
In [2]: struct.unpack("<i", b"xaax00x00xaa")
Out[2]: (-1442840406,)
In [3]: struct.unpack("<I", b"xaax00x00xaa")
Out[3]: (2852126890,)
I would also like a similar solution to work for int16 and uint16, but first things first, since I guess an extension would be trivial if I manage to solve this problem.
c++ int
c++ int
asked Nov 24 '18 at 19:38
user10668188user10668188
193
asked Nov 24 '18 at 19:38
user10668188user10668188
193
asked Nov 24 '18 at 19:38
user10668188user10668188
193
asked Nov 24 '18 at 19:38
user10668188user10668188
193
asked Nov 24 '18 at 19:38
user10668188user10668188
193
193
2
Because you're placing them all in (signed)intvariables.
– Rotem
Nov 24 '18 at 19:45
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49
add a comment |
2
Because you're placing them all in (signed)intvariables.
– Rotem
Nov 24 '18 at 19:45
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49
2
2
Because you're placing them all in (signed)
int variables.– Rotem
Nov 24 '18 at 19:45
Because you're placing them all in (signed)
int variables.– Rotem
Nov 24 '18 at 19:45
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
You need to store the unsigned values in an unsigned variables and it will work:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
unsigned int v = *(unsigned int *) a.c_str();
int32_t x = *(int32_t *) a.c_str();
uint32_t y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
When you cast the value to unsigned and then store it in a signed variable, the compiler plays along. Later, when you print the signed variable, the compiler generates code to print a signed variable output.
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to store the unsigned values in an unsigned variables and it will work:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
unsigned int v = *(unsigned int *) a.c_str();
int32_t x = *(int32_t *) a.c_str();
uint32_t y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
When you cast the value to unsigned and then store it in a signed variable, the compiler plays along. Later, when you print the signed variable, the compiler generates code to print a signed variable output.
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
add a comment |
You need to store the unsigned values in an unsigned variables and it will work:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
unsigned int v = *(unsigned int *) a.c_str();
int32_t x = *(int32_t *) a.c_str();
uint32_t y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
When you cast the value to unsigned and then store it in a signed variable, the compiler plays along. Later, when you print the signed variable, the compiler generates code to print a signed variable output.
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
add a comment |
You need to store the unsigned values in an unsigned variables and it will work:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
unsigned int v = *(unsigned int *) a.c_str();
int32_t x = *(int32_t *) a.c_str();
uint32_t y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
When you cast the value to unsigned and then store it in a signed variable, the compiler plays along. Later, when you print the signed variable, the compiler generates code to print a signed variable output.
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
You need to store the unsigned values in an unsigned variables and it will work:
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string a("xaax00x00xaa", 4);
int u = *(int *) a.c_str();
unsigned int v = *(unsigned int *) a.c_str();
int32_t x = *(int32_t *) a.c_str();
uint32_t y = *(uint32_t *) a.c_str();
std::cout << a << std::endl;
std::cout << "int: " << u << std::endl;
std::cout << "uint: " << v << std::endl;
std::cout << "int32_t: " << x << std::endl;
std::cout << "uint32_t: " << y << std::endl;
return 0;
}
When you cast the value to unsigned and then store it in a signed variable, the compiler plays along. Later, when you print the signed variable, the compiler generates code to print a signed variable output.
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
answered Nov 24 '18 at 19:46
John MurrayJohn Murray
869514
869514
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
add a comment |
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
Great! The biggest risk of being lazy and just copy-paste rows... Thanks!
– user10668188
Nov 24 '18 at 20:45
add a comment |
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2
Because you're placing them all in (signed)
intvariables.– Rotem
Nov 24 '18 at 19:45
Ahhh! How stupid of me... Been pulling my hair for hours. Thanks.
– user10668188
Nov 24 '18 at 19:49