Say I have a binary file; it contains positive binary numbers, but written in little endian as 32-bit integers
How do I read this file? I have this right now.
int main() {
FILE * fp;
char buffer[4];
int num = 0;
fp=fopen("file.txt","rb");
while ( fread(&buffer, 1, 4,fp) != 0) {
// I think buffer should be 32 bit integer I read,
// how can I let num equal to 32 bit little endian integer?
}
// Say I just want to get the sum of all these binary little endian integers,
// is there an another way to make read and get sum faster since it's all
// binary, shouldnt it be faster if i just add in binary? not sure..
return 0;
}
In the case of little endian format, the least significant byte appears first, followed by the most significant byte. The letter 'T' has a value of 0x54 and is represented in 16 bit little endian as 54 00.
Big endian machine: An int is 4 bytes, and the first is the largest. I read 4 bytes (W X Y Z) and W is the largest. The number is 0x12345678. Little endian machine: Sure, an int is 4 bytes, but the first is smallest.
If machine is little endian then *c will be 1 (because last byte is stored first) and if the machine is big endian then *c will be 0. Does endianness matter for programmers?
Although many processors use little-endian storage for all types of data (integer, floating point), there are a number of hardware architectures where floating-point numbers are represented in big-endian form while integers are represented in little-endian form.
This is one way to do it that works on either big-endian or little-endian architectures:
int main() {
unsigned char bytes[4];
int sum = 0;
FILE *fp=fopen("file.txt","rb");
while ( fread(bytes, 4, 1,fp) != 0) {
sum += bytes[0] | (bytes[1]<<8) | (bytes[2]<<16) | (bytes[3]<<24);
}
return 0;
}
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