Question

C language problem.

Suppose I open I file and read a integer 24. And I want to store them into a array byte [].

The answer should be byte[0] = 0x9F.

How can I do that?

Answer 1

#include <stdio.h>

int main(void) {
        int x = 24;

        // in C, there is no byte type, however char is of 1
        // byte, hence we can treat it same way.
        unsigned char byteArr[4];

        byteArr[0] = x & 0xFF;
        x = x >> 8; // shift to right by 8 bits.
        
        byteArr[1] = x & 0xFF;
        x = x >> 8; // shift to right by 8 bits.
        
        byteArr[2] = x & 0xFF;
        x = x >> 8; // shift to right by 8 bits.
        
        byteArr[3] = x & 0xFF;
        
        printf("0th byte(hex): %X\n", byteArr[0]);
        printf("1st byte(hex): %X\n", byteArr[1]);
        printf("2nd byte(hex): %X\n", byteArr[2]);
        printf("3rd byte(hex): %X\n", byteArr[3]);
}

**************************************************
0x9F is equal to 16*9 + 15 = 159 in decimal.. So Your answer is incorrect. 24 is equal to 18 => 16 + 8 = 24 in decimal.


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