函数的隐式声明 [重复]

Question

:

warning: implicit declaration of function (9 answers)

Closed 11 months ago .

First of all I have to said that my brain is burned because of this question. I need to write a function that checks whether the 3rd parameter is a interleaved of 1st and 2nd parameter. I wrote it but when I run it, it gives me this error. "warning: implicit declaration of function 'interleaved'" Here is my code:

void main()
{
    char str1[100] = "ABC";
    char str2[100] = "XYZ";
    char str3[100] = "XAYZBC";

    printf(interleaved(str1, str2, str3));
}

bool interleaved(char str1[], char str2[], char str3[])
{
    int i = 0, j = 0, k = 0;

    if (strlen(str1) + strlen(str2) != strlen(str3))
        return false;

    while (k < strlen(str3))
    {
        if (i < strlen(str1) && str1[i] == str3[k])
            i++;

        else if (j < strlen(str2) && str2[j] == str3[k])
            j++;
        else
            return false;

        k++;
    }

    if (!(i == strlen(str1) && j == strlen(str2) && k == strlen(str3)))
        return false;

    return true;
}

I found the faulty part and tried to fix it as follows:

void main()
{
    char str1[100] = "ABC";
    char str2[100] = "XYZ";
    char str3[100] = "XAYZBC";

    bool result = interleaved(str1[100], str2[100], str3[100]);
    printf(result);
}

Can anyone help me please ?

Answer 1

"Implicit declaration" usually means, at the time when you are trying to use (call) a function, the compiler does not know the function signature / prototype, hence cannot determine the return type and cannot do any check on the argument passing.

Most of the time this is solved by either

• Defining the function before it's use
• Forward declaring the function.

That said, your function call and usage of printf() is not proper. You can do something like

bool result = interleaved(str1[100], str2[100], str3[100]);
printf("%s", result?"yes":"no");

Which prints "yes" is the function returns true, and "no" otherwise.

Answer 2

You have to declare or define functions to use before using them.

Also bool is not a correct data to pass as the first argument of printf() .

Try this:

bool interleaved(char str1[], char str2[], char str3[]); /* declare the function */

int main(void) /* use standard type for main() */
{
    char str1[100] = "ABC";
    char str2[100] = "XYZ";
    char str3[100] = "XAYZBC";

    printf("%s\n", interleaved(str1, str2, str3) ? "true" : "false"); /* correctly use printf() */
}

Answer 3

You are trying to use the result of interleaved (which is a bool ) as the format string for printf , which is not allowed.

Instead, you need to do something like this:

printf("%d", interleaved(str1, str2, str3));

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Also, you need to at least declare interleaved before you make a call to that function.