can 'fast' type prevent conversion from int to uint?

Question

Consider the following program:

#include <stdio.h>
#include <stdint.h>

int main()
{
    uint16_t result;
    uint16_t ui = 1;
    int16_t si = -1;
    
    result = si * ui;
    
    printf("%i", result);

    return 0;
}

This prints the value 65535, which is what I expect after having read this post: si is converted to ui, so max+1 is added to it. Now, in the next code snippet, I change the type of result to uint_fast16_t.

#include <stdio.h>
#include <stdint.h>

int main()
{
    uint_fast16_t result;
    uint16_t ui = 1;
    int16_t si = -1;
    
    result = si * ui;
    
    printf("%li", result);

    return 0;
}

Now, the result is -1. What happens here? How can the result be signed?

Answer 1

Please see below code(As @Tom Karzes said that uint_fast16_t may be an ordinary unsigned int on some systems. In that case, %lu would be the wrong format. And @bolov said that should use printf("%" PRIuFAST16 "\n", result); and printf("%" PRIdFAST16 "\n", (int_fast16_t) result); So I changed my answer):

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>


int main()
{
    uint_fast16_t result;
    uint16_t ui = 1;
    int16_t si = -1;
    
    result = si * ui;
    printf("sizeof(result) is %zu\n", sizeof(result));

    printf("%" PRIuFAST16 "\n", result);
    printf("%" PRIdFAST16 "\n", (int_fast16_t) result);

    return 0;
}

Run it will output(on 64-bit computer):

sizeof(result) is 8
18446744073709551615
-1

Why output is different for the same result? One is %PRIuFAST16, other is %PRIdFAST16.

It's because it depend on how you see it, as unsigned or signed.