How to swap string in C? - c

Question

I am trying to make a function in C which will swap two string variables but something went wrong and the program crashes.
Please have a look at my code and tell me where I made a mistake:
#include <string.h>
void strswap(char name1[], char name2[]) // to swap two strings
{
int lengthname1, lengthname2;
lengthname1 = strlen(name1);
lengthname2 = strlen(name2);
char temporaryname1[100];
char temporaryname2[100];
int x;
int y;
// till just the declaration
for (int x = 0; x < lengthname1; lengthname1++) {
temporaryname1[x] = name1[x];
name1[x] = ' ';
}
// copying the value of name1 in temporaryname1
for (int y = 0; y < lengthname2; lengthname2++) {
temporaryname2[x] = name2[x];
name2[x] = ' ';
}
// copying the value of name2 in temporaryname2
for (int x = 0; x < lengthname1; lengthname1++) {
name1[x] = temporaryname2[x];
}
for (int y = 0; y < lengthname2; lengthname2++) {
name2[x] = temporaryname1[x];
}
}
#include <stdio.h>
int main()
{
char name[] = "hello";
char name2[] = "hi";
printf("before swapping: %s %s\n", name, name2);
strswap(name, name2);
printf("after swapping: %s %s\n", name, name2);
}
EDIT:- I have corrected the program and it is working properly. Soon my header file is going to work with some other modules. Thank you all for your help and especially to #Micheal

Answer 1

There are many issues:
First issue
The x variable is not initialized:
int x; int y; // first declaration of x
// till just the declaration
for(int x=0;x<lengthname1;lengthname1++)
{// ^ second declaration of x , local to the loop
temporaryname1[x]=name1[x];
name1[x]=' ';
}
// if you use x here it's the first x that has never been initialized
Second issue
This:
for (x = 0; x<lengthname1; lengthname1++)
should be:
for (x = 0; x<lengthname1 + 1; x++)
Why lengthname1 + 1 ? Because you need to copy the NUL char that terminates the string.
There are similar problems in your other for loops as well.
For example here you use y as loop variable, but in the loop you use x:
for (int y = 0; y<lengthname2 + 1; lengthname2++)
{
name2[x] = temporaryname1[x];
Third issue
In main you declare this:
char name[] = "hello";
char name2[] = "hi";
That is actually the same as
char name[6] = "hello"; // 5 chars for "hello" + 1 char for the terminating NUL
char name2[3] = "hi"; // 2 chars for "hi" + 1 char for the terminating NUL
Now even if your strswap is correct, you are trying to stuff the 6 bytes from the name array ("hello") into the 3 bytes array name2, there is not enough space in the name2 array. This is undefined behaviour.
And last but not least:
This is simply useless:
name1[x] = ' ';
And finally
You should ask yourself why you need two temporary strings (temporaryname1 and temporaryname2) in strswap() - one is enough.

Answer 2

void strswap(char ** name1, char ** name2)
{
char * name1_1 = malloc(strlen(*name2) + 1);
char * name2_1 = malloc(strlen(*name1) + 1);
strncpy(name1_1, *name2, strlen(*name2) + 1);
strncpy(name2_1, *name1, strlen(*name1) + 1);
*name1 = name1_1;
*name2 = name2_1;
}
int main()
{
char * name="hello";
char * name2="hi";
printf("before swapping %s %s\n",name,name2);
strswap(&name, &name2);
printf("after swapping %s %s\n",name,name2);
return 0;
}
Actually following is the safest way to swap two strings. In your cases, you were statically using strings of size 100 in size, which is not good and work in all case. Moreover, You tried to use ' ' instead of '\0' to mark end of string. String APIs uses '\0' to indicate that string has ended.