How To Replace A Certain Character in C - c

Question

I am writing a program that replaces characters in the user's input in C but I don't know how to replace the certain characters. Is there a certain method for C that replaces characters in a string? If you know python, then I want something a bit like this in python:
string.replace('certain_character','replacement_character')
Something like that, except for C, and not python.
This is my code that I have written so far:
#include <stdio.h>
int main(){
char str[BUFSIZ];
printf("Welcome To My Secret Language encoder!Enter some text: \n");
scanf("%s",str);
/*
Where I want to replace certain characters
*/
printf("Here is your text in secret language mode: %s \n",str);
}
I'm writing this code to learn C more, and that's why i'm not doing it in a higher level language like python.So, how do you replace certain characters in a string?

Answer 1

Nothing like that in C. You'll have to scan the string yourself:
#include <string.h>
char str[] = "I love cats";
int i;
for(i = 0; i < strlen(str); i++)
{
if(str[i] == 'c')
str[i] = 'b';
}
Now, if you're looking for a substring, you'll need something like strstr.

Answer 2

strchr finds the given character in a string, or returns NULL.
int main() {
int c;
while ( ( c = getchar() ) != EOF ) {
char const * found, * source = "abc", * dest = "xyz";
if ( ( found = strchr( "abc", c ) ) != NULL ) {
putchar( dest[ found - source ] );
} else {
putchar( c );
}
}
return 0;
}

Answer 3

If you have a lot of characters that you want to replace with other characters (like a Caesar cypher) you can build a lookup for yourself as follows:
#include <string.h>
char plain[] = "Hello there good people";
char encoder[26] = "ghijklmnopqrstuvwxyzabcdef";
char secret[100]; // long enough
int n = strlen(plain);
for(ii = 0; ii < n; ++ii) {
secret[ii] = encoder[(tolower(plain[ii]) - 'a')%26];
}
secret[n] = '\0';
This uses a couple of tricks:
cast all characters to lower case
subtract 'a' from the lowercase number - since a char is really just a number, we now have a == 0
Perform a modulo operation on the result so things that fall outside of the range of good characters don't cause a memory access error.
Add a '\0' at the end to make sure the string is properly terminated.
Copying things into a new string; obviously you could do an in-place replacement.
As written this will turn numbers (digits) and punctuation / symbols / spaces into characters. You could decide that anything that is not a letter is maintained - and maybe that only lower case letters are converted. In that case
#include <string.h>
char plain[] = "Hello there good people";
char encoder[26] = "ghijklmnopqrstuvwxyzabcdef";
char secret[100]; // long enough
int n = strlen(plain);
for(ii = 0; ii < n; ++ii) {
if(plain[ii] >= 'a' && plain[ii] <= 'z') {
secret[ii] = encoder[plain[ii] - 'a'];
}
else {
secret[ii] = plain[ii];
}
}
secret[n] = '\0';

Answer 4

there is no such function, you have to write one using strstr.
if you can use std::string, you can use string.replace()

Answer 5

Say you want to replace: A with z and b with X
char *replace(char *src, int replaceme, int newchar)
{
int len=strlen(src);
char *p;
for(p=src; *p ; p++)
{
if(*p==replaceme)
*p=newchar;
}
return src;
}
usage:
replace(string, 'A', 'z');
replace(string, 'b', 'X');
This is just the logic to do it, you need more statements in your code.