How to change arrays using void functions in C? - c

Question

I just started trying to learn C a couple days ago by following along with the K&R book. I am not totally new to programming but I would say I am a novice. I've been doing mostly fine so far but I've been having a lot of trouble with array manipulation. In section 2.8 of the book, it defines a function called squeeze which takes in an array of characters and a single character and removes all instances of the single character from the array. I was confused at first because this was a void function.
#include <stdio.h>
void squeeze(char s[], char c);
int main(){
char s[] = "hello";
squeeze(s, 'l');
printf("%s\n", s);
}
void squeeze(char s[], char c){
int i, j;
for (i = j = 0; s[i] != '\0'; i++){
if (s[i] != c)
s[j] = s[i];
++j;
}
s[j] = '\0';
}
I implemented it into my own C program and, when I call the function in Main, it just prints my original array without altering it. I am reasonably sure that this is because I am not changing the array s[] with the Squeeze function before I call printf, but rather changing a copy of the array that isn't being used. I have researched using pointers to refer to arrays in C but I am still very confused on how to do that correctly, because no matter what I try I either get a segment error or the array is not changed. Can anyone help me understand how pointers and arrays work in this context?
EDIT: I did originally state the name of the parameter in the function prototype and declaration, but I made an error in copying it.

Answer 1

I am reasonably sure that this is because I am not changing the array s[] with the Squeeze function
Function prototype is ok:
void squeeze(char [], char c);
It's totally ok to parse array here, because array will decay to pointer anyway in the context of function parameter.
Your implementation wasn't correct, though:
You must provide the name of the parameter s
And must enclose your if statement properly, otherwise ++j; will be executed no matter whether the condition if (s[i] != c) is met or not.
Here's a guide to fix the two issues:
void squeeze(char s[], char c){ // <-- 1) must provide parameter name
int i, j;
for (i = j = 0; s[i] != '\0'; i++){
if (s[i] != c)
{ // <-- 2) must enclose if statement properly
s[j] = s[i];
++j;
}
}
s[j] = '\0';
}

Answer 2

you are changing the array with the squeeze function. But the squeeze function cant produce the output you needed beacause there is an implementation problem with the function.
#include <stdio.h>
void squeeze(char [], char c);
int main(){
char s[] = "hello";
squeeze(s, 'l');
printf("%s\n", s);
}
void squeeze(char s[], char c){
int i, j;
for (i = j = 0; s[i] != '\0'; i++){
if (s[i] != c)
s[j++] = s[i];
}
s[j] = '\0';
}
This one will produce the output you needed. Because in the case of if statement If you not gave any curly brackets The immediate line to the if loop will be taken as in the loop.
So in the case of
if (s[i] != c)
s[j] = s[i];
++j;
The ++j wont taken as in the loop. If you want to add ++j to loop. Then use the following code.
if (s[i] != c){
s[j] = s[i];
++j;
}

Answer 3

All are false in the internet， you should reassign the new variable
void squeeze(char s[],char c) {
printf("s=%s\n",s);
int i, j;
char new_s[20];
for (i = j = 0; s[i] != '\0'; i++)
{
if (s[i] != c)
{
new_s[j] = s[i];
j++; //这里加了一但是 s[j] 没有赋值，下一个循环时才赋值。。
//而最后一次没有下一个循环，所以s[j] 还没有赋值
}
}
printf("s=%s\n",new_s);
}