[ACCEPTED]-How do I set the number of characters output in a fprintf '%s' format using a variable?-printf

Accepted answer
Score: 19

This one corresponds to your example:

fprintf(file, "%*s", 10, string);

but 2 you mentioned a maximum as well, to also 1 limit the number:

fprintf(file, "%*.*s", 10, 10, string);
Score: 1

I believe you need "%*s" and you'll need 2 to pass the length as an integer before 1 the string.

Score: 1

As an alternative, why not try this:

void print_limit(char *string, size_t num)
  char c = string[num];
  string[num] = 0;
  fputs(string, file);
  string[num] = c;

Temporarily 4 truncates the string to the length you want 3 and then restores it. Sure, it's not strictly 2 necessary, but it works, and it's quite 1 easy to understand.

Score: 0

As an alternative, you may create "c_str" from 4 your buffer and prepare string to printf 3 like I do that in this source:

void print(char *val, size_t size) {
    char *p = (char *)malloc(size+1); // +1 to zero char
    char current;
    for (int i=0; i < size; ++i) {
        current = val[i];
        if (current != '\0') {
           p[i] = current; 
        } else {
           p[i] = '.'; // keep in mind that \0 was replace by dot
        p[i+1] = '\0';
    printf("%s", p);

But it solution 2 wrong way. You shuld use fprintf with format 1 "%*s".

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