Replacing a character from a string in C

Question

I tried with this method but the output remains the same as the input. The user enters the character he wants to replace and with which letter he wants to replace with. I don't understand where did I go wrong.

#include<stdio.h>

char* replaceChar(char *s, char x,char y)
{
    int i=0;
    while(s[i])
    {
        if(s[i]==x)
        {
            s[i]==y;
        }
        i++;
    }

    return s;

}

int main()
{
    char string[30];

    printf("Enter the string:\n");
    gets(string);
    fflush(stdin);
    char x;
    char y;
    printf("Enter the character you want to replace:\n");
    scanf("%c",&x);
    printf("Enter the character you want to replace with:\n");
    scanf(" ");
    scanf("%c",&y);

    printf("After replacing the string :\n");
    printf("%s",replaceChar(&string[0],x,y));


    return 0;
}

Obligatory: [Never use `gets()`](https://stackoverflow.com/q/1694036/10077) — Fred Larson, Jan 20 '21 at 14:52
Yes, use `fgets()`. And be aware that it leaves the `\n` in the resulting string, which `gets()` does not. — Fred Larson, Jan 20 '21 at 15:01
It's always helpful to include sample input and expected and actual output. — Caleb, Jan 20 '21 at 15:03

Vlad from Moscow · Accepted Answer · 2021-01-20T15:01:08.230

The problem is that instead of the assignment operator you are using the comparison operator in this code snippet

    if(s[i]==x)
    {
        s[i] == y;
    }

Write

    if(s[i]==x)
    {
        s[i] = y;
    }

Pay attention to that the function gets is unsafe and is not supported by the C Standard any more. Instead use the function fgets.

Also this call

fflush(stdin);

has undefined behavior. Remove it.

And use

scanf(" %c",&x);
      ^^^

scanf(" %c",&y);
      ^^^

instead of

scanf("%c",&x);
scanf(" ");
scanf("%c",&y);

Replacing a character from a string in C

1 Answers1