String made NUL after function cal - c

Question

I am trying to recreate the C++ function std::string.pop_back in C, where the last non-NUL character of the input string is popped off. In the pop_back function below the variable out has the desired output of the function, but trying to reassign *string to be out makes *string be the correct value before the function closes (for example, if I were to print string* at the end of the function, it would output Hello World), but in main(), after pop_back() is called, string is output as blank. What am I doing wrong in reassigning string?
#include <stdio.h>
#include <string.h>
void pop_back(char** string) {
size_t stringLen = strlen(*string);
char out[stringLen];
strxfrm(out, *string, stringLen);
// Here *string == "Hello World!"
*string = out;
// Here *string == "Hello World"
}
int main(int argc, char* argv[]) {
char* string = "Hello World!";
printf("Initial string: %s\n", string);
pop_back(&string);
printf("After pop_back: %s\n", string);
return 0;
}
// Output:
// $ ./pop_back_test
// Initial string: Hello World!
// After pop_back:
// Expected output:
// $ ./pop_back_test
// Initial string: Hello World!
// After pop_back: Hello World

Answer 1

With your code
char out[stringLen];
...
*string = out;
You "return" (assign to a parameter) a pointer to a "local variable", i.e. a pointer to an object with automatic storage duration, which's life time ends at the end of the function. Accessing an object beyond its lifetime is undefined behaviour.
You probably have to dynamically allocate a new string, which you can then manipulate and use after function execution has ended. For example:
if (!(*string) || !(**string)) {
// decide what to do with NULL or empty input strings.
...
}
else {
size_t stringLen = strlen(*string);
char* out = malloc(stringLen);
memcpy(out, *string, stringLen);
out[stringLen-1] = '\0';
}
*string = out;
Don't forget to free allocated memory afterwards. BTW: if you operate on a copy, I'd suggest to return the (new) copy and leave the input pointer as is, i.e. I'd change the prototype to
char* pop_back(char* string) {
...
return out;
}

Answer 2

As I step through your code, the message in my debugger indicates on line...
*string = out;
...that *string is over array bounds, failing the assignment. *string points to a string literal which means among other things it cannot be modified. Attempting to modify its content will always fail.
There are several ways this can be remedied.
One way is to modify your pop_back() function to accept a char *, and return a char *, allowing you to pass in an immutable string and return the resulting string.
Suggested implementation:
char * pop_back(char *string)
{
size_t stringLen = strlen(string)+1;//add one for NULL
char *out = calloc(stringLen,1);//create space for new string
strxfrm(out, string, stringLen);
// Here *string == "Hello World!"
return out;//return string
// Here *string == "Hello World"
}
int main(int argc, char* argv[])
{
char *string1 = "Hello World!"; // source string
char *string2 = {0}; // container for local result
printf("Initial string: %s\n", string1);
string2 = pop_back(string1); // get local result
printf("After pop_back: %s\n", string2);// use local result
free(string2); // free local result
return 0;
}