looking for efficient swap char pointer

Question

My question is about efficiency of the swap char pointer algorithm.

Why either of the commented lines causes my complier to throw error?

The same logic works for swapping b with temp, but the same logic causes error while repeating in the same function for the temp and a?

Thank you all.

#include <iostream>
void sawp ( char *a, char *b, int l)
{
    char * prtA = a;
    char * prtB = b;
    char *temp = (char *)calloc(l, sizeof(char));

    while(*(char*)b)
    {        
        *(char*)temp= *(char*)b;
        // *(char*)b  = *(char*)a;
        // *(char*)a = *(char*)temp;

        std::cout<<*(char*)temp<<std::endl;
        b++;
        temp++;
        a++;    
    }

}

int main()

{
    char *a="Name";
    char *b="Lastname";
    sawp (a,b,sizeof(b));
    return 0;

}

Side note #1: Those `(char*)` castings are completely redundant - you are casting a `char*` to a `char*`. Side note #2: You've misspelled `swap` as `sawp`. Side note #3: You're not doing anything with `prtA` and `prtB`. Side note #4: You've misspelled "compiler" as "complier". All these side notes are just to tell you that you should put a little effort before posting a question here (that is, if you expect others to put an effort answering it). Important note #1: It's not the compiler that "throws an error", because it is not a compilation error - it's a runtime error!!! — barak manos, Dec 10 '14 at 18:35

Gopi · Accepted Answer · 2014-12-10T18:35:34.650

2

Modifying string literals results in undefined behavior. It might cause a crash. String literals are read only and hence they can't be modified.

So

*b = *a;

is an error.

If you want to modify your string then they should have read and write access. So

char a[10] = "name1";
char b[10] = "name2";

Then you can have

*b = *a; /* In your sawp() function */

edited Dec 10 '14 at 18:35

answered Dec 10 '14 at 18:14

Gopi

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score 2 · Answer 2 · answered Dec 11 '14 at 20:28

Writing char *a="Name" the compiler will treat the "Name" string as a constant string and will place it in the rom memory section. The rom memory section is a read only section and the code has no permission to change it in run time, so it will rise a segmentation fault when you try to change it.

score 0 · Answer 3 · edited May 23 '17 at 12:21

0

Did you took a look at this entry? Swapping pointers in C (char, int)

In the end, there's an explanation on how to swap chars. Maybe it fits you.

edited May 23 '17 at 12:21

Community

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answered Dec 10 '14 at 18:15

pedrop

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score 0 · Answer 4 · answered Dec 10 '14 at 18:21

To avoid the issue with literals as posted by Gopi, you could use character arrays instead of pointers in main, and make sure they are the same size, for example:

char a[10] = {'N','a','m','e',0,0,0,0,0,0};
char b[10] = {'L','a','s','t','n','a','m','e',0,0};
/* ... */
    sawp(a, b, sizeof(b));

or if you still wanted to use pointers, you could use malloc() (assuming this is C and not C++) to allocate space for the strings, and then copy literals into the strings you allocated.

looking for efficient swap char pointer

4 Answers4