使用写入 function 重新编码 printf %p，没有 printf

Question

I am currently working on a task where I need to print the address of a variable. It would be easy to use printf %p but I am only allowed to use write from unistd.

I tried casting the pointer in to an unsigned integer and uintptr_t and then converting it into a hexadecimal number. With uintptr_t it works but with an unsigned integer it only prints half of the address. Maybe someone can explain me why this is the case?

I also saw some solutions using ">>" and "<<" but I didn't get why that works. It would be nice if someone can explain a solution using "<<" and ">>" step by step, because I am not sure if I am allowed to use uintptr_t.

this is the code I use to cast it into a unsigned int / unitptr_t / unsigned long long (I know that ft_rec_hex is missing leading 0's):

void ft_rec_hex(unsigned long long nbr)
{
    char tmp;

    if (nbr != 0)
    {
        ft_rec_hex(nbr / 16);
        if (nbr % 16 < 10)
            tmp = nbr % 16 + '0';
        else
            tmp = (nbr % 16) - 10 + 'a';
        write(1, &tmp, 1);
    }
}

int     main(void)
{
    char c = 'd';
    unsigned  long long ui = (unsigned long long)&c;
    ft_rec_hex(ui);
}

Answer 1

It looks like only half of the address is printed because the "unsigned integer" you used has only half size of uintptr_t . (note that uintptr_t is an unsigned integer type)

You can use an array of unsigned char to store data in a pointer variable and print that to print full pointer withput uintptr_t .

Using character types to read objects with other type is allowed according to strict aliasing rule .

#include <stdio.h>
#include <unistd.h>

void printOne(unsigned char v) {
    const char* chars = "0123456789ABCDEF";
    char data[2];
    data[0] = chars[(v >> 4) & 0xf];
    data[1] = chars[v & 0xf];
    write(1, data, 2);
}

int main(void) {
    int a;
    int* p = &a;
    /* to make sure the value is correct */
    printf("p = %p\n", (void*)p);
    fflush(stdout);

    unsigned char ptrData[sizeof(int*)];
    for(size_t i = 0; i < sizeof(int*); i++) {
        ptrData[i] = ((unsigned char*)&p)[i];
    }
    /* print in reversed order, assuming little endian */
    for (size_t i = sizeof(int*); i > 0; i--) {
        printOne(ptrData[i - 1]);
    }

    return 0;
}

Or read data in a pointer variable as unsigned char array without copying:

#include <stdio.h>
#include <unistd.h>

void printOne(unsigned char v) {
    const char* chars = "0123456789ABCDEF";
    char data[2];
    data[0] = chars[(v >> 4) & 0xf];
    data[1] = chars[v & 0xf];
    write(1, data, 2);
}

int main(void) {
    int a;
    int* p = &a;
    /* to make sure the value is correct */
    printf("p = %p\n", (void*)p);
    fflush(stdout);

    /* print in reversed order, assuming little endian */
    for (size_t i = sizeof(int*); i > 0; i--) {
        printOne(((unsigned char*)&p)[i - 1]);
    }

    return 0;
}

Answer 2

It would be easy to use printf %p but I am only allowed to use write from unistd.

Then form a string and print that.

int n = snprintf(NULL, 0, "%p", (void *) p);
char buf[n+1];
snprintf(buf, sizeof buf, "%p", (void *) p);
write(1, buf, n);

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Using a pointer converted to an integer marginally reduces portability and does not certainly form the best textual representation of the pointer - something implementation dependent.

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With uintptr_t it works but with an unsigned integer it only prints half of the address.

unsigned is not specified to be wide enough to contain all the information in a pointer.

uintptr_t , when available (very common), can preserve most of that information for void pointers. Good enough to round-trip to an equivalent pointer, even if in another form.