Trim whitespaces

Hi,

I am trying to write a C program to trim whitespaces before the string. But, it isn't working properly. Following is what I tried

char *lefttrim(char *str)
{
int i=0;
while(strlen(str) > 0 && str[0] == ' ')
{
for(i = 1; i< strlen(str); i++)
{
str[i-1] = str[i];
str[strlen(str)] = '\0';
}
return(str);
}

It would be great if someone can point out some mistake.

Thanks,
Pingoo

str is a local variable to your function you need to dereference it for it to have effect outside the function. Returning it is bad since it will be popped of the stack once the function returns.

How are you using the function?

I am taking cmd line arguments and calling this function to trim the whitespaces.

int main(int argc, char **argv)
{
char str[100];
int i;

for(i=1; i<=strlen(str); i++)
{
lefttrim(argv[i]);
printf("The value of i is %d\n", i);
printf("Trimed string is %s\n",argv[i]);
}
exit(0);

http://cboard.cprogramming.com/c-pro...ead-first.html
Read this before posting any more code.

Actually forget what I said before, you are dereferencing it. But you need to change the index where your terminating zero is placed. Although returning a pointer to the string seems pointless since you don't use it and it's actually pointing the the base of the string you are sending the function.

Quote:

---

Originally Posted by Pingoo

I am taking cmd line arguments and calling this function to trim the whitespaces.

int main(int argc, char **argv)
{
char str[100];
int i;

for(i=1; i<=strlen(str); i++)
{
lefttrim(argv[i]);
printf("The value of i is %d\n", i);
printf("Trimed string is %s\n",argv[i]);
}
exit(0);

---

And what do you expect str to contain in this example? Also, what does str have to do with argv[i]?

Also it's helpful if you can describe the problem beyond "it's not working". In this case I think it would be nice if you could provide, what the program does, and how that differs from what you are expecting.

Code:

---

char *lefttrim(char *str)
{
    int i=0;
    while(strlen(str) > 0 && str[0] == ' ')
    {
        for(i = 1; i< strlen(str); i++)
        {
            str[i-1] = str[i];
            str[strlen(str)] = '\0';
        }
        return(str);
    }

---

Do you see error now?

After doing the required changes, I am getting following output

$ gcc -o trim trim_whitespace.c
$ ./trim This is a test string.
The value of i is 1
Trimed string is This
The value of i is 2
Trimed string is is
The value of i is 3
Trimed string is a
The value of i is 4
Trimed string is test
The value of i is 5
Trimed string is string.
Segmentation fault
---------------------------------------
I don't know why it is saying Segmantation fault.

What condition are you using instead of strlen(str) as claudiu pointed out in your for loop?

As you can see, the input is 5 strings. The segmentation fault happens after string 5, it appears as you are doing operations on argv[6].

Looks like you should have access to "gdb" for debugging if you are on linux, which makes it easy to find seg faults. Have a look here, it will take you ~5 min:

http://cboard.cprogramming.com/c-pro...tml#post938474

Worth noting as well is that your test strings doesn't contains any space, so you aren't really testing your function. You would need to enter them like this:

Code:

---

./trim "  this" "  is" "  a" "  test."

---

Thanks for correcting me. It is working fine now.

gdb was useful too.

Good to hear that. What is your current code? I noticed that you used strlen() in a loop, but you actually do not need to do that.