Printing one extra empty line at the beginning of output in C

Question

Why is this printing one empty line at the beginning of the output? \n is only after %s... Help please, I'm so frustrated.

if(argc > 1){

    while(r!=NULL){

        r = fgets(str, MAXL, stdin);

        if(r==NULL){
            return 0;
        }

        if (*argv[1] == 'i'){
            char *invP = inv(r);
            printf("%s\n", invP);
            free(invP);
        }

inv() is:

char* inv(char* C){
int length = 0;

int i;
for(i = 0; C[i]!='\0'; i++){
    length++;
}

char *inverted;
inverted = malloc(length+1);
inverted[length] = '\0';
char* invP = inverted;

int j = 0;
for(i = length - 1; i >= 0; i--){
    inverted[j] = C[i];
    j++;
}
return invP;
}

It doesn't have any print on it, dunno why is this happening.

You're going to need to show us more information. What does your `inv` function do, for example? — Xymostech, Mar 24 '13 at 21:36
@IlyaBoltnev No, because `inv` creates a new block of memory with `malloc` for its return value. — Xymostech, Mar 24 '13 at 21:41

score 7 · Accepted Answer · edited May 23 '17 at 12:28

7

When you call fgets, your r string has a newline at the end. So, when you invert it (by calling inv), the string ends up with the newline at the beginning. Then, when you print it, you see the newline first.

r -> "hello\n"
invP = "\nolleh"

If you want to remove the newline, you can use something like

char *pos;
if ((pos=strchr(r, '\n')) != NULL)
    *pos = '\0';

(taken from here)

edited May 23 '17 at 12:28

Community

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answered Mar 24 '13 at 21:39

Xymostech

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Xymostech You're ABSOLUTELY right! I can't use strlib in this project, so I fixed it in my inv() method by switching for(i = 0; C[i]!='\0'; i++){ for for(i = 0; C[i]!='\n'; i++){ Thank you very much I was going crazy!! – Punk Mar 24 '13 at 21:50

Printing one extra empty line at the beginning of output in C

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