Setting a buffer of char* with intermediate casting to int*

Question

I could not fully understand the consequences of what I read here: Casting an int pointer to a char ptr and vice versa

In short, would this work?

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Answer 1

This code might be of help to you. It shows a 32-bit number being built by assigning its contents a byte at a time, forcing misalignment. It compiles and works on my machine.

#include
#include
#include
#include

int main () {
    uint32_t *data = (uint32_t*)malloc(sizeof(uint32_t)*2);
    char *buf = (char*)data;
    uintptr_t addr = (uintptr_t)buf;
    int i,j;
    i = !(addr%4) ? 1 : 0;
    uint32_t x = (1<<6)-1;
    for( j=0;j<4;j++ ) buf[i+j] = ((char*)&x)[j];

    printf("%" PRIu32 "\n",*((uint32_t*) (addr+i)) );
}

As mentioned by @Learner, endianness must be obeyed. The code above is not portable and would break on a big endian machine.

Note that my compiler throws the error "cast from ‘char*’ to ‘unsigned int’ loses precision [-fpermissive]" when trying to cast a char* to an unsigned int, as done in the original post. This post explains that uintptr_t should be used instead.