How to read hex values into integer with fstream (C++)
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0
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I am trying to read a little-endian hex string from a binary file, and put that value into an integer to work with it. When I try to read, instead of getting a number I get ascii symbols. I've tried casts and atoi and nothing seems to work. What is the best way to use fstream to read a hex string into an integer from a file?
This is essentially my program:
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main(int argc, char* argv) {
fstream input;
fstream output;
char cbuffer[4];
char revbuffer[8];
input.open(argv[1], fstream::binary | fstream::in);
output.open("output.txt", ios::out | ios::app);
input.seekg(16, input.beg);
input.read(cbuffer, 4);
cout << sizeof(revbuffer) << endl;
cout << cbuffer[0] << cbuffer[1] << cbuffer[2] << cbuffer[3] << endl;
}
c++ input hex fstream atoi
add a comment |
up vote
0
down vote
favorite
I am trying to read a little-endian hex string from a binary file, and put that value into an integer to work with it. When I try to read, instead of getting a number I get ascii symbols. I've tried casts and atoi and nothing seems to work. What is the best way to use fstream to read a hex string into an integer from a file?
This is essentially my program:
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main(int argc, char* argv) {
fstream input;
fstream output;
char cbuffer[4];
char revbuffer[8];
input.open(argv[1], fstream::binary | fstream::in);
output.open("output.txt", ios::out | ios::app);
input.seekg(16, input.beg);
input.read(cbuffer, 4);
cout << sizeof(revbuffer) << endl;
cout << cbuffer[0] << cbuffer[1] << cbuffer[2] << cbuffer[3] << endl;
}
c++ input hex fstream atoi
is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to read a little-endian hex string from a binary file, and put that value into an integer to work with it. When I try to read, instead of getting a number I get ascii symbols. I've tried casts and atoi and nothing seems to work. What is the best way to use fstream to read a hex string into an integer from a file?
This is essentially my program:
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main(int argc, char* argv) {
fstream input;
fstream output;
char cbuffer[4];
char revbuffer[8];
input.open(argv[1], fstream::binary | fstream::in);
output.open("output.txt", ios::out | ios::app);
input.seekg(16, input.beg);
input.read(cbuffer, 4);
cout << sizeof(revbuffer) << endl;
cout << cbuffer[0] << cbuffer[1] << cbuffer[2] << cbuffer[3] << endl;
}
c++ input hex fstream atoi
I am trying to read a little-endian hex string from a binary file, and put that value into an integer to work with it. When I try to read, instead of getting a number I get ascii symbols. I've tried casts and atoi and nothing seems to work. What is the best way to use fstream to read a hex string into an integer from a file?
This is essentially my program:
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main(int argc, char* argv) {
fstream input;
fstream output;
char cbuffer[4];
char revbuffer[8];
input.open(argv[1], fstream::binary | fstream::in);
output.open("output.txt", ios::out | ios::app);
input.seekg(16, input.beg);
input.read(cbuffer, 4);
cout << sizeof(revbuffer) << endl;
cout << cbuffer[0] << cbuffer[1] << cbuffer[2] << cbuffer[3] << endl;
}
c++ input hex fstream atoi
c++ input hex fstream atoi
edited Nov 19 at 9:42
dwalter
5,38912233
5,38912233
asked Nov 19 at 9:16
Kasper
31
31
is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41
add a comment |
is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41
is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
If it's an integer value stored in binary format, I guess it's either a int32_t
or a uint32_t
. Since you mention that the value is stored in little-endian byte order, I guess you want to make sure that the host running your program converts it (if it needs to). C++20 has std::endian. If that's not available to you, there are usually macros for detecting endianness at compiletime that you can use instead of the std::endian
tests I've used. I've assumed that the value is a uint32_t
below.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <type_traits> // std::endian
// little endian unsigned 32 bit integer to host byte order
inline uint32_t Le32toh(uint32_t le) {
#if __cplusplus <= 201703L
// run-time check
static constexpr uint16_t endian = 1;
if(*reinterpret_cast<const uint8_t*>(&endian)==1) return le;
#else
// compile-time check
static_assert(std::endian::native == std::endian::little || std::endian::native == std::endian::big);
if constexpr (std::endian::native == std::endian::little) return le;
#endif
const uint8_t* c=reinterpret_cast<const uint8_t*>(&le);
return // little-to-big endian conversion
(static_cast<uint32_t>(c[0])<<24) |
(static_cast<uint32_t>(c[1])<<16) |
(static_cast<uint32_t>(c[2])<<8) |
(static_cast<uint32_t>(c[3]));
return le;
}
int main(int argc, char* argv) {
std::vector<std::string> args(argv+1, argv+argc);
std::fstream output("output.txt", std::ios::out | std::ios::app);
uint32_t cbuffer;
for(const auto& file : args) {
std::fstream input(file, std::fstream::binary | std::fstream::in);
input.seekg(16, input.beg);
// read directly into the varibles memory
input.read(reinterpret_cast<char*>(&cbuffer), 4);
// output the value unconverted
std::cout << std::hex << cbuffer << "n";
// convert if needed
cbuffer = Le32toh(cbuffer);
// output the value converted
std::cout << std::hex << cbuffer << "n";
}
}
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.
– Ted Lyngmo
Nov 20 at 0:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If it's an integer value stored in binary format, I guess it's either a int32_t
or a uint32_t
. Since you mention that the value is stored in little-endian byte order, I guess you want to make sure that the host running your program converts it (if it needs to). C++20 has std::endian. If that's not available to you, there are usually macros for detecting endianness at compiletime that you can use instead of the std::endian
tests I've used. I've assumed that the value is a uint32_t
below.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <type_traits> // std::endian
// little endian unsigned 32 bit integer to host byte order
inline uint32_t Le32toh(uint32_t le) {
#if __cplusplus <= 201703L
// run-time check
static constexpr uint16_t endian = 1;
if(*reinterpret_cast<const uint8_t*>(&endian)==1) return le;
#else
// compile-time check
static_assert(std::endian::native == std::endian::little || std::endian::native == std::endian::big);
if constexpr (std::endian::native == std::endian::little) return le;
#endif
const uint8_t* c=reinterpret_cast<const uint8_t*>(&le);
return // little-to-big endian conversion
(static_cast<uint32_t>(c[0])<<24) |
(static_cast<uint32_t>(c[1])<<16) |
(static_cast<uint32_t>(c[2])<<8) |
(static_cast<uint32_t>(c[3]));
return le;
}
int main(int argc, char* argv) {
std::vector<std::string> args(argv+1, argv+argc);
std::fstream output("output.txt", std::ios::out | std::ios::app);
uint32_t cbuffer;
for(const auto& file : args) {
std::fstream input(file, std::fstream::binary | std::fstream::in);
input.seekg(16, input.beg);
// read directly into the varibles memory
input.read(reinterpret_cast<char*>(&cbuffer), 4);
// output the value unconverted
std::cout << std::hex << cbuffer << "n";
// convert if needed
cbuffer = Le32toh(cbuffer);
// output the value converted
std::cout << std::hex << cbuffer << "n";
}
}
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.
– Ted Lyngmo
Nov 20 at 0:20
add a comment |
up vote
0
down vote
accepted
If it's an integer value stored in binary format, I guess it's either a int32_t
or a uint32_t
. Since you mention that the value is stored in little-endian byte order, I guess you want to make sure that the host running your program converts it (if it needs to). C++20 has std::endian. If that's not available to you, there are usually macros for detecting endianness at compiletime that you can use instead of the std::endian
tests I've used. I've assumed that the value is a uint32_t
below.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <type_traits> // std::endian
// little endian unsigned 32 bit integer to host byte order
inline uint32_t Le32toh(uint32_t le) {
#if __cplusplus <= 201703L
// run-time check
static constexpr uint16_t endian = 1;
if(*reinterpret_cast<const uint8_t*>(&endian)==1) return le;
#else
// compile-time check
static_assert(std::endian::native == std::endian::little || std::endian::native == std::endian::big);
if constexpr (std::endian::native == std::endian::little) return le;
#endif
const uint8_t* c=reinterpret_cast<const uint8_t*>(&le);
return // little-to-big endian conversion
(static_cast<uint32_t>(c[0])<<24) |
(static_cast<uint32_t>(c[1])<<16) |
(static_cast<uint32_t>(c[2])<<8) |
(static_cast<uint32_t>(c[3]));
return le;
}
int main(int argc, char* argv) {
std::vector<std::string> args(argv+1, argv+argc);
std::fstream output("output.txt", std::ios::out | std::ios::app);
uint32_t cbuffer;
for(const auto& file : args) {
std::fstream input(file, std::fstream::binary | std::fstream::in);
input.seekg(16, input.beg);
// read directly into the varibles memory
input.read(reinterpret_cast<char*>(&cbuffer), 4);
// output the value unconverted
std::cout << std::hex << cbuffer << "n";
// convert if needed
cbuffer = Le32toh(cbuffer);
// output the value converted
std::cout << std::hex << cbuffer << "n";
}
}
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.
– Ted Lyngmo
Nov 20 at 0:20
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If it's an integer value stored in binary format, I guess it's either a int32_t
or a uint32_t
. Since you mention that the value is stored in little-endian byte order, I guess you want to make sure that the host running your program converts it (if it needs to). C++20 has std::endian. If that's not available to you, there are usually macros for detecting endianness at compiletime that you can use instead of the std::endian
tests I've used. I've assumed that the value is a uint32_t
below.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <type_traits> // std::endian
// little endian unsigned 32 bit integer to host byte order
inline uint32_t Le32toh(uint32_t le) {
#if __cplusplus <= 201703L
// run-time check
static constexpr uint16_t endian = 1;
if(*reinterpret_cast<const uint8_t*>(&endian)==1) return le;
#else
// compile-time check
static_assert(std::endian::native == std::endian::little || std::endian::native == std::endian::big);
if constexpr (std::endian::native == std::endian::little) return le;
#endif
const uint8_t* c=reinterpret_cast<const uint8_t*>(&le);
return // little-to-big endian conversion
(static_cast<uint32_t>(c[0])<<24) |
(static_cast<uint32_t>(c[1])<<16) |
(static_cast<uint32_t>(c[2])<<8) |
(static_cast<uint32_t>(c[3]));
return le;
}
int main(int argc, char* argv) {
std::vector<std::string> args(argv+1, argv+argc);
std::fstream output("output.txt", std::ios::out | std::ios::app);
uint32_t cbuffer;
for(const auto& file : args) {
std::fstream input(file, std::fstream::binary | std::fstream::in);
input.seekg(16, input.beg);
// read directly into the varibles memory
input.read(reinterpret_cast<char*>(&cbuffer), 4);
// output the value unconverted
std::cout << std::hex << cbuffer << "n";
// convert if needed
cbuffer = Le32toh(cbuffer);
// output the value converted
std::cout << std::hex << cbuffer << "n";
}
}
If it's an integer value stored in binary format, I guess it's either a int32_t
or a uint32_t
. Since you mention that the value is stored in little-endian byte order, I guess you want to make sure that the host running your program converts it (if it needs to). C++20 has std::endian. If that's not available to you, there are usually macros for detecting endianness at compiletime that you can use instead of the std::endian
tests I've used. I've assumed that the value is a uint32_t
below.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <type_traits> // std::endian
// little endian unsigned 32 bit integer to host byte order
inline uint32_t Le32toh(uint32_t le) {
#if __cplusplus <= 201703L
// run-time check
static constexpr uint16_t endian = 1;
if(*reinterpret_cast<const uint8_t*>(&endian)==1) return le;
#else
// compile-time check
static_assert(std::endian::native == std::endian::little || std::endian::native == std::endian::big);
if constexpr (std::endian::native == std::endian::little) return le;
#endif
const uint8_t* c=reinterpret_cast<const uint8_t*>(&le);
return // little-to-big endian conversion
(static_cast<uint32_t>(c[0])<<24) |
(static_cast<uint32_t>(c[1])<<16) |
(static_cast<uint32_t>(c[2])<<8) |
(static_cast<uint32_t>(c[3]));
return le;
}
int main(int argc, char* argv) {
std::vector<std::string> args(argv+1, argv+argc);
std::fstream output("output.txt", std::ios::out | std::ios::app);
uint32_t cbuffer;
for(const auto& file : args) {
std::fstream input(file, std::fstream::binary | std::fstream::in);
input.seekg(16, input.beg);
// read directly into the varibles memory
input.read(reinterpret_cast<char*>(&cbuffer), 4);
// output the value unconverted
std::cout << std::hex << cbuffer << "n";
// convert if needed
cbuffer = Le32toh(cbuffer);
// output the value converted
std::cout << std::hex << cbuffer << "n";
}
}
edited Nov 20 at 0:21
answered Nov 19 at 9:34
Ted Lyngmo
1,202213
1,202213
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.
– Ted Lyngmo
Nov 20 at 0:20
add a comment |
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.
– Ted Lyngmo
Nov 20 at 0:20
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Thanks for the reply. I'm getting an error: 'std::endian' has not been declared. But with main alone I'm now able to read the value as an int. Thanks very much!
– Kasper
Nov 19 at 22:54
Great! Yes,
std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.– Ted Lyngmo
Nov 20 at 0:20
Great! Yes,
std::endian
is a C++20 feature so to be able to do something similar during compile time in C++17 and earlier you'd have to resort to some non-standard endian macro - or do a runtime check. I'll add that too.– Ted Lyngmo
Nov 20 at 0:20
add a comment |
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is the value really stored as a hex string in the file? You mentioned the byte order and you read 4 chars from the file so a hex string seems improbable.
– Ted Lyngmo
Nov 19 at 11:39
Maybe not then, I don't really know how to call it. I'm fairly new to programming. The value is at offset 0x10 and looks something like [in hex] 00 92 f8 03. The number should read 0x03f89200.
– Kasper
Nov 19 at 22:41