C language replaces spaces in strings with%20

subject ： Please implement a function , Replace each space in the string with "%20". Examples ： “abc defgx yz” turn “abc%20defgx%20yz”

First time to see this topic , I wrote the following error code , First of all, let's think about what's wrong ？

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h> 
#include<assert.h>
  char *replace(char *p)
{
    
	assert(p != NULL);
	char *s = p;
	while (*p)
	{
    
		if (*p == ' ')
			*p = "%20"; 
		p++;
 	}
	return s;
}
int main()
{
    
	char a[] = "abc defgx yz";
	replace(a);

	printf("%s", a);// The result is ：abc 躣躣躣躣躣躣躣躣躣躣躣 efgx 躽 z

	return 0;
}

In fact, the simple idea of the above code is through pointers , Traverse this character array , If you find a space, replace it with %20, Always find the end of the string \0 Up to . But for this problem, I have neglected an important problem , Is that we find the space and replace it with %20, A space is a character ,%20 It's three characters , Replace three characters with one character , How can memory space be filled , Causes the array to be out of bounds , As for the result, it is also strange .

An array can't be troublesome , To find a space, you need to move the following element , Then we can define two arrays

#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
#include<string.h>
void replace(char *arr1, char *arr2)
{
    
	while (*arr1)
	{
    
		if (*arr1 != ' ')
		{
    
			*arr2 = *arr1;
			arr2++;
		}
		else
		{
    
			strcpy(arr2, "%20");
			arr2 = arr2 + 3;
		}
		arr1++;
	}
	*arr2 = '\0';

}
int main()
{
    
	char arr1[] = "abc defgx yz";;// The array size is 13
	char arr2[40];// Arbitrarily define the appropriate array size , I can put it down arr1 The content of 
	replace(arr1, arr2);
	printf("%s", arr2);
	return 0;
}

Program run results ：

The basic implementation idea of the above code is ：

• Define two arrays ,arr1 Used to store the original string ,arr2 Used to store added %20 String after
• Traverse the array through the pointer arr1, Look for spaces , Until you find the string \0 until , If it is not a space during traversal , Character by character arr1 Put the content of arr2 Inside , If you encounter a space, put it after the character that is not a space , hold %20 Copy to the end of the string ,arr2 The space in front is full , So the position of the next character should be moved back three positions . Continue to implement this Law .
• Finally, when you find arr1 Of \0 Replacement complete , stay arr2 After the replaced character, add \0 that will do （ because arr2 Only defined but not initialized , There's nothing in it ）

Another algorithm is also ok , Look at the code first ：

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include<string.h>

char * replace (char *str)
{
    
	char *p = str;
	char arr[30];
	while (*p != '\0')
	{
    
		if (*p == ' ')
		{
    
			strcpy(arr, p + 1);
			*p = '%';
			*(p + 1) = '2';
			*(p + 2) = '0';
			strcpy(p + 3, arr);
			p = p + 3;// Three positions have been occupied ,p Move back three positions 
			continue;//p Already moved , So don't move , Use continue Skip the following code p++
		}
		p++;
	}
	return str;
}
int main()
{
    
	char arr1[30] = "abc defgx yz";//14
	//char arr2[40];
	replace(arr1);
	printf("%s", replace(arr1));
	return 0;
}

This algorithm idea is a little similar to the above , The code above is added to the second array %20 after , Output this array arr2, Do not use the original array ; The following idea is to borrow an array to temporarily store things , I added %20 after , Then put the things temporarily stored behind the blank , Get back the original array .

The basic idea of the above code is ： Every time you need to save the string after the space into an array , Then replace the space with %20 Then copy the string just copied to %20 You can just .