C - No output data

Question

I have a problem with my source code. When I do this, I have no output data, why ? How I can resolve this please ?
Thank's !

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

void ft_split_whitespaces(char *str);

int main(int argc, char** argv)
{
    ft_split_whitespaces("Hello\nThis is\ta\nWord");
    return (0);
}

void ft_split_whitespaces(char *str)
{
    int i;

    for (i = 0; str[i] != '\0'; ++i);
    {
        if (str[i] == '\n' || str[i] == '\t' || str[i] == ' ')
        {
            printf("Ascii table (int): %d, Ascii table (char): %c\n", str[i], str[i]);
        }
    }
}

Answer 1

Typographical error - semicolon at the end of for means empty statement - for (i = 0; str[i] != '\0'; ++i);.

Fix: remove semicolon and never put one even instead of empty statement at the end of for/while/if - use { /* intentionally empty */} instead

for (i = 0; str[i] != '\0'; ++i) // no semicolon here ever.
{
    if (str[i] == '\n' || str[i] == '\t' || str[i] == ' ')
    {
        printf("Ascii table (int): %d, Ascii table (char): %c\n", str[i], str[i]);
    }
}

Answer 2

What Alexei said.

To avoid that, I like to use this sort of curly bracing:

for (i = 0; str[i] != '\0'; ++i){
    if (str[i] == '\n' || str[i] == '\t' || str[i] == ' '){
        printf("Ascii table (int): %d, Ascii table (char): %c\n", str[i], str[i]);
    }
}

This way the first thing you do after completing the conditions in a loop is put an open curly brace, instead of accidentally put a semicolon by reflex, and then jump down and put an open curly brace.