Thread: vsprintf ommit literal percent character

Hello, recently I have had a problem while trying to makeing a function that sets some formated text to a Status control on a window based application, the problem is basically that vsprintf is ommiting any percentaje (%) character that I want to use as a literal character (for example, while searching something I need to inform the user in a way like that: "Reached 25% of files"); first I thought that was a window thing, so I tested in console mode and I get the same result. That's the testing code I used:

Code:

#include <stdio.h>
#include <stdarg.h>


void apply_text(char *data, ...) {
    
    va_list args;
    char output_text_buffer[100];
    
    if(data) {
        va_start(args, data);
        vsprintf(output_text_buffer, data, args);
        va_end(args);
        
        //testing result
        printf("%s", output_text_buffer);
    }
}


int main() {
    
    apply_text("Reach 25% of files");
    
    getchar();
    return 0;
}

Even if I use a backslash before the %, or if I use a %c to print the ASCII 37 it is ommiting the percentaje, the result is "Reach 25 of files".

How can I solve that?
Thank's in advance

Niara

% is the escape character for all the print type functions. Notice how you use "%s" to print a string. You need to use two percent signs to get a literal percent "... 25%% ...".

Yeah it works, thanks both anduril462 and oogabooga for your responses, I can go on now.