Sending a pointer of a character array to a function in C - c

Question

Okay, long story short I am having some trouble:
I'm trying to send the starting address of a character array (string) to a function. Once the function gets the pointer, I'd like the function to parse through the characters one by one (in order to modify certain characters). As of right now I just set it up to print each character, but I'm still failing at that really badly.
This is what I have:
#include "system.h"
#include <stdio.h>
typedef unsigned int uint32;
typedef unsigned short uint16;
typedef unsigned char uint8;
void EXCLAIM(char *msg){
char the_char = 0;
the_char = msg;
while (the_char != 0)
{
printf(the_char);
*msg = *(msg++);
the_char = *msg;
}
}
int main(void) {
char *first_str = "This is a test. Will this work. I. am. Not. Sure...";
while (1) {
EXCLAIM(first_str);
}
}
EDIT:
Here is the updated code with what I was trying to do; send the pointer and go through each character replacing all periods with exclamation marks.
#include "system.h"
#include <stdio.h>
typedef unsigned int uint32;
typedef unsigned short uint16;
typedef unsigned char uint8;
void exclaim(char *msg){
int i;
for( i=0; msg[i]; i++ )
{
if (msg[i] == '.') {
msg[i] = '!';
}
}
printf(msg);
}
int main(void) {
char *the_sentences = "This is a test. Will this work. I. am. Not. Sure...";
while (1) {
exclaim(the_sentences);
}
}
Thank you all for your help!

Answer 1

Your main() is fine.
The issue lies in the pointer arithmetics inside EXCLAIM.
I can see two different versions that are useful:
Using pointer arithmetics
while (*msg)
{
printf("%c", *msg);
msg++
}
Using indexing
int i;
char the_char;
for( i=0; msg[i]; i++ )
{
printf( "%c", msg[i] );
}

Answer 2

Try something like this:
#include <stdio.h>
void exclaim(char * s)
{
while (*s != '\0')
{
putc(*s);
++s;
}
}
Notes:
Don't shout. Don't use all caps for anything but preprocessor macros.
Print characters, not strings.
No need to make another copy. The function argument is already a local variable that you can use directly.

Answer 3

You should use:
the_char = *msg;
not:
the_char = msg;

Answer 4

*msg = *(msg++);
What are you trying to do with this line of code? Do you understand what it actually does? First off, you need to know that msg++ changes the value of the pointer. Then you assign whatever it USED to point to to the new location. This means that the whole string will be replaced with copies of the first character in the string.
The moral of the story is to not do too much in a single line of code. You have to be especially careful with the increment operator ++. Even seasoned programmers typically put something like msg++ in its own line because it gets too complicated when you try to mix it in with a complex expression.

Answer 5

Something that was not mentioned but, attemping to modify a string literal is undefined behavior.
You need to change the following line from,
char *the_sentences = "This is a test. Will this work. I. am. Not. Sure...";
to an array.
char the_sentences[] = "This is a test. Will this work. I. am. Not. Sure...";
Then feel free to modify the contents inside your function.
#include <stdio.h>
void foo(char *msg)
{
for (; *msg; msg++)
if (*msg == '?')
*msg = '!';
}
int main(void)
{
char line[] = "Hello, world?? How are you tonight??";
foo(line);
puts(line);
return 0;
}