How can I make a char* point to a char[]?

Question

I was wondering if there is a way to make char* point to the contents of a char array so that I can modify the char* across functions.

For example

void toup(char* c) {
  char array[sizeof(c)];
  for (int x;x<strlen(c);x++){
    array[x]=toupper(c[x]);
  }
}

int main(){
  char *c="Hello";
  toup(c);
}

Trying to make the array = char* does not seem to work. Is it possible to make the char* point to the char array?

Answer 1

Is it possible to make the char* point to the char array?

Yes. Instead of:

int main(){
  char *c="Hello";
  toup(c);
}

Use:

int main(){
  char c[] = "Hello";
  toup(c);
}

char *c = "Hello"; makes the string const and usually puts the string in a const data section. char c[] = "Hello"; provides the mutable string you want.

Also see Why is conversion from string constant to 'char*' valid in C but invalid in C++.

---

Also see Blaze's comment:

for (int x;x<strlen(c);x++) x is uninitialized. Did you mean int x = 0?

---

Two other caveats...

void toup(char* c) {
  char array[sizeof(c)];
  for (int x;x<strlen(c);x++){
    array[x]=toupper(c[x]);
  }
}

First, toup is modifying a local array. It is not visible outside the function.

Second, sizeof(c) yields either 4 or 8 since it is taking the size of the pointer. That means the declaration is either char array[4]; on 32-bit machines, or char array[8]; on 64-bit machines.

array[x]=toupper(c[x]); should segfault when the length of string c is larger then the pointer.

You should probably do something like:

void toup(char* c) {
  for (size_t x=0;x<strlen(c);x++){
    c[x]=toupper(c[x]);
  }
}

A similar question is at How to iterate over a string in C? Also see What is array decaying?

Answer 2

No need for temporary buffer array - you already have stream of characters in input.

#include <stdio.h>
#include <ctype.h>

void toup(char* c) {
  for (char * it = c; *it !='\0'; it++){
    *it = toupper(*it);
  }
}

int main(){
  char c[] = "Hello";
  toup(c);
  printf("%s\n",c);
}