How to replace strings in C, just using stdio.h??? Help, please!!!

Question

i need to create a small program: user enter 1st string ex: I'm a beautiful girl then he needs to replace for example 'beautiful' to 'clever'

can use just stdio.h and pointers as well

help me, pls!!!

Answer 1

You should probably take a look at the fgets function of stdio.h

Answer 2

Here is the outline of an algorithm

1. Parse arguments search-string and replacement-string
2. Read input line
3. Search for search-string in input line and replace it by replacement-string
4. Output line

Step 3 will have to use handcrafted loops for string operations because the typically used functions are found in <string.h>

Answer 3

First, write the code and use functions in string.h as usual. Now you know which functions you need to replace.

Then, research and/or ask specific questions to do with replacing those functions.

Hope this helps.

Answer 4

#include <stdio.h>
#define SIZE 1000

void read_str();


main() {
    char search_key;
    char *ptr1;

    char s1[SIZE];

    ptr1 = s1;


    printf("Enter your string:\n");
    read_str();

    printf("Enter the key word u want to change:");
    scanf("%c", &search_key);
    // after this i don't have idea what to do....((((((

    return 0;
}

void read_str()
{
    int c;
    while((c = getc(stdin))!= '\n')
    *ptr1++ = c;
    *ptr1++ = ' ';
    *ptr1 = '\0';
}

Answer 5

Not sure about C but I did something similar in C++. C might have a similar range of string manipulations. userInput="I am a beautiful girl" replaceString=userInput.replace(5,9,"Clever");

Answer 6

First, let us suppose you want just literal substitution and not something more complicated that would distinguish from "beautifulgirl" and "beautiful girl".

First, write your compare function.

• Inputs: ptr to buffer (buf), word to find (w)
• (1) while ( *w != '\0')
• if *buf != *w return 0 (false)
• buf++, w++ and loop (1)
• return 1 (true)

Then scan your input buf and copy it to dbuf (they must be big enough);

• i = 0; j = 0;
• (1) if ( buf[i] == '\0' ) dbuf[j] = 0; exit loop
• if compare(&buf[i], "beautiful") is 1 (true), copy "clever" to &dbuf[j], then skip "beautiful" (update i) and update j (for "clever") and loop (1)
• dbuf[j] = buf[i]; i++; j++; loop (1)

This will change every instance of "beautiful", being a word or not.

To skip beautiful, it is enough to increment i by the length of beautiful; the dest buffer index j must be instead incremented by the length of "clever". To avoid hardencoding lengths, the following

int strlen(char *s)
{
  char *p = s;
  while( *s != '\0' ) s++;
  return s-p;
}

should be a possible impl of strlen (change its name eventually).

Bugfixes and real code implementation are left to you (since the task is very boring indeed)