Are there good tutorial for advanced programming c?

That's only a snippet of large code where I am swapping the two array elements which are strings so the problem is certainly not that

fixed a little. should work now

swap(&arr[0], &arr[1]);

man that just worked

remember that if the type doesn't match then this is UB. converting a char (*)[10] to const char * for example.

But how?

What does UB means?

#include <stdio.h> void swap(const char **lhs, const char **rhs) { const char *temp = *lhs; *lhs = *rhs; *rhs = temp; } int main() { const char *arr[] = {"Hello", "World!"}; puts(arr[0]); puts(arr[1]); swap(&arr[0], &arr[1]); puts(arr[0]); puts(arr[1]); return 0; }

in the swap function replace const char * with int and this is an ordinary pointer based swap

i did't get the syntax you used int the function decleartion can you tell me?

in the swap function replace const char * with int and this is an ordinary pointer based swap

and do this

i did't get the syntax you used int the function decleartion can you tell me?

void swap(int *lhs, int *rhs) { int temp = *lhs; *lhs = *rhs; *rhs = temp; }

void swap(int *lhs, int *rhs) { int temp = *lhs; *lhs = *rhs; *rhs = temp; }

😱 man I am getting crazy

oof. in that case you need to read a bit about pointers

Initially it was char * and I passed argument arr[0] which is also a char *

ye. and you tried to replace the char contents pointed to by the char *

But when you make it const char ** and I passed &arr[0] which is address of char *

i did not replace the contents of the string, i replaced the contents of your array

But content of the array are characters

in this case the array has const char * members. but if you do char arr[3][4] for example, the members are of type char[4]. conversion from char (*)[4] to const char * and then accessing the data using that pointer is undefined behaviour

But content of the array are characters

i and j are indexes

like this swapval(arr[i], arr[j]);

Isn't this syntax tells you I and j are indexes of same arr array

what is the type of the array?

Array of strings

oof. in that case you need to read a bit about pointers

You asked me? Go back and show me the message

arr is the array i just told you. if that am saying am giving array element as the argument why don't you believe?

You didn't even asked me which type of array it is ? Show me when you said which type of array. And you are free to go

the question is implicit in what is arr

From where?

like look at my answers. every time i mentioned an array i either mentioned its type or the member type @Devil420H

Prata. just understanding my int example should be enough.

Prata. just understanding my int example should be enough.

And not getting your example that's why am asking

then you can replace any type for int and do the same thing

But am not getting how it's happening

;-; Prata = C Primer Plus by Stephen Prata