CODEWITHSUNDEEP
cloops
Bloodknofsky
Bloodknofsky

Reputation: 21

I am getting this warning "comparison of constant ‘10’ with boolean expression is always true"

I am unable to escape from the loop by entering newline.

int input(char str[], int n)
{
    int ch, i = 0;
    while ((ch == getchar()) != '\n')
        if (i < n)
            str[i++] = ch;
    str[i] = '\0';
    return i;
}

Upvotes: 1

Views: 551

Answers (3)

Henry Le Berre
Henry Le Berre

Reputation: 910

Just replace == with = because you want to assign getchar() to ch and not compare it.

int input(char str[], int n)
{
    int ch, i = 0;
    while ((ch = getchar()) != '\n'){
        if (i < n)
            str[i++] = ch;
    }
    str[i] = '\0';
    return i;
}

Upvotes: 4

RobertS supports Monica Cellio
RobertS supports Monica Cellio

Reputation: 15042

(ch == getchar())

You use == instead of =.

== is used as boolean operator for checking if the left hand operand equals right hand one.

Thus, the loop will never terminate because you compare 0 or 1 (which is the result of (ch == getchar()) with \n whose ASCII value is 10.

That´s what the error is saying.

= is used for assignment and required to assign the character fetched by getchar() to ch.

This is the corrected version:

while ((ch = getchar()) != '\n') {
     if (i < n)
     str[i++] = ch;
}

Upvotes: 0

Welgriv
Welgriv

Reputation: 823

Yo forgot { symbols in your code. Any control structure such while and if of more than one instruction line should be coded like this :

while(condition){
  instruction 1;
  instruction 2;
  ...
}

By the way your i is always 0 and your n does not change within your function.

EDIT: The reason you obtain your warning have been answered by someone else.

Upvotes: 0

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