CODEWITHSUNDEEP
cvisual-studio
Yellowfun
Yellowfun

Reputation: 558

Differences in compilers. C language

Why my "C" code doesn't compile in Linux (compiler ggdb3, C99) but nice work in Visual Studio? Here message of error:

20:1: error: control may reach end of non-void function [-Werror,-Wreturn-type] }

#include <stdio.h>
//  Function to determine if one character string exists inside another string.
int findString(char source[], char searching[])
{
    int i = 0, j = 0;

    for (; source[i] != '\0'; i++) {
        if (source[i] == searching[j])
            if (source[i + 1] == searching[j + 1] &&
                source[i + 2] == searching[j + 2])
                return i;
            else
                return -1;
    }

}

int main(void)
{
    int index = findString("a chatterbox", "hat");

    if (index == -1)
        printf("No match are founded\n");
    else
        printf("Match are founded starting index is %i\n", index);

    return 0;
}

I've tried to edd in the function this but it doesn't help

if (source[0] == '\0')
        return -1;

Upvotes: 0

Views: 89

Answers (2)

tdao
tdao

Reputation: 17678

First C is not Python so you need to use brackets properly (not just indenting).

That said, the problem is with your findString() function. Once you do put some brackets properly, you will that if (source[i] != searching[j]) the function does not have a return statement - whereas it's expected to return an int.

 if (source[i] == searching[j])
 {
     ...
 }
 // what if source[i] != searching[j]
 // you do not have any return statement for a function returning int

Failure to return from a non-void function causes undefined behaviour.

Citing C11 (by rubenvb)

6.9.1 Function definitions

12 If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

It's more clear with C++:

[...]Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function.[...]

Upvotes: 1

user2933559
user2933559

Reputation:

It looks like you're actually just getting a warning, but since you've given the Linux compiler the command line argument -Werror it's treating the warning as an error. If you look in the compiler output for Visual Studio you should see a similar warning.

Upvotes: 2

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