Thread: function return problem

Code:

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

char* reverse(char* a)
{
  int temp, x, y = strlen(a) - 1;

  for ( x = 0; x < y; x++, y-- )
    temp = a[x], a[x] = a[y], a[y] = temp;

  return a;
}

int main()
{
    char *text = "hello world";

How do I reverse variable text using reverse() and assign the result to a new variable char*? Thanks.

temp = a[x], a[x] = a[y], a[y] = temp;

These are separate statements, need to separated by ;

When you post a program like this, it's always a good idea to state what specifically is the problem. For example, any errors messages or unexpected behavior.

I would imagine any problems are coming from your use of compound assignment statements. Try and separate them out a bit to make them easier for you to read.

Everything works fine. You don't need to worry about function reverse(). My problem is that I don't know how to assign what reverse() returns to a new variable.

Code:

y = f(x);

Where y is the variable you want the return value assigned to, f is your function, and x is your parameter.

OK, but what type should y be?

Whatever type you're returning. In this case, a pointer to char (char *).

It doesn't work

Code:

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

char* reverse(char* a)
{
  int temp, x, y = strlen(a) - 1;

  for ( x = 0; x < y; x++, y-- )
    temp = a[x], a[x] = a[y], a[y] = temp;

  return a;
}

int main()
{
	char *text = "hello world";
	char *reversed = reverse(text);

	return 0;
}

Try to run that.

Everything works fine

Apparently not. Try making the changes already suggested to you. On second inspection, your return function is in serious need of a complete re-write. You take a pointer to an array as a parameter, but then treat that pointer like the array itself.

You are trying to modify a string literal. String literals are allowed to be placed in read only memory.

Try changing

Code:

char *text

to

Code:

char text[]

Worked
Thanks!