How can I create a long int from a char buffer?

user_12460366 0
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I have a char buffer with 4 numbers, in binary:

buffer[0] = 00000000
buffer[1] = 00000000
buffer[2] = 00000000
buffer[3] = 00000001

I want to "fetch" the 4 bytes of the buffer and create a long int.

In this case, the long int variable would be:

test == 00000000 00000000 00000000 00000001

This is my code but I have the feeling that it is conceptually wrong:

int main() {
         unsigned char buffer [4];
         buffer[0] = 0;
         buffer[1] = 0;
         buffer[2] = 0;
         buffer[3] = 1;
         for (int i = 0; i < 4; i++)
                 printf ("%d\n", buffer[i]);

         long int * addr = (long int *) &buffer[0];
         long int test = * addr;
         printf ("test = %li\n", test);
}

Output:

0
0
0
1
test = 140728915197952

I was trying to use pointers because I think it's the right way but I don't understand myself at all.

My machine is LittleEndian if can help.

long-integer c arrays
2 Answers

You can use left shift with bitwise or in the loop, like this:

uint32_t l = 0;
for(int i=0;i<4;i++){
    l<<=8;
    l|=buffer[i];
}
printf("l=%x\n",l);

Shift and Bitwise operations are typically very fast in all processors.

Suggestion: int, long may have different size. To ensure 32bit size, please use uint32_t instead.

user_117747430

The code could be:

#include <inttypes.h>
// ...

uint32_t test = buffer[0] * 0x1000000u + buffer[1] * 0x10000u + buffer[2] * 0x100u + buffer[3];
printf("test = " PRIu32 "\n", test);

It would be a good idea to use a fixed width type for this scenario , because unsigned long might have different sizes on different platforms .

user_15059390
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