在 C 中重用格式说明符

Question

Is there a way to reuse a format specifier in C? I would like to use the same specifier for the next 20 outputs, is there an easier way than to write the specifier 20 times? v

In Fortran I can do this by writing 20 in front of the specifier, so (20F10.0) would be the same as (F10.0 F10.0 F10.0...). Is there something similar in C?

Answer 1

At the risk of your preprocessor macros looking like a snake with a severe stutter and your future self wondering why on earth you've done something like this, you could (ab)use the fact that string constants can be concatenated...

#include <stdio.h>

#define REPEAT5(s) s s s s s 
#define REPEAT4(s) s s s s
#define REPEAT20(s) REPEAT4(REPEAT5(s))

int main() {
    printf(REPEAT20("%d "), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20);
}

If you run this on Godbolt.org you can see the format string was correctly expanded to

.asciz  "%d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d "

Answer 2

int x;
for(x=0;x<20;x++)
printf("%d", x);




It's just a diagram. In the 
body of the for loop, you 
can independently determine 
what exactly printf ( )will print. 

Answer 3

This should do the trick:

char *n_format(char *format, size_t n, char delimiter)
{
    char *result;
    int  index = 0;
    size_t len;

    if (!format)
        return (NULL);
    len = strlen(format);
    if (!(result = malloc(len * n + n)))
        return (NULL);
    for (int i = 0; i < n; ++i)
    {
        strcpy(result + index, format);
        index += len;
        result[index++] = delimiter;
    }
    result[index - 1] = '\0';
    return (result);
}

Usage:

int main()
{
    char *string;
    string = n_format("%.5f", 5, ' ');
    printf("|%s|\n", string);
    // output: |%.5f %.5f %.5f %.5f %.5f|
    free(string);
    return (0);
}