功能指针-变量未初始化

Question

#include <stdio.h>

int Calc(int n1, int n2, int(*func) (int n1, int n2))
{
    return func(n1, n2);
}

int Plus(int n1, int n2)
{
    return n1 + n2;
}

int Minus(int n1, int n2)
{
    return n1 - n2;
}

int main(void)
{
    int n1, n2;
    int result;
    char mark;

    printf("입력 : ");
    scanf_s("%d", &n1);
    printf("choose function : ");
    scanf_s("%c", &mark, sizeof(mark));
    printf("입력 : ");
    scanf_s("%d", &n2);

    switch (mark)
    {
    case '+':
        result = Calc(n1, n2, Plus);
    case '-':
        result = Calc(n1, n2, Minus);
    }
    printf("%d\n", result);

    return 0;
}

I`m studying Function Pointer in C with simple code. I made it initialize variable result from using scanf_s, but it send me a error message that inform me variable result is not initialized. How can I fix this code?

Answer 1

You have two problems:

1. you leave '\\n' in the input buffer after scanf_s("%d", &n1); which is then taken as your input to scanf_s("%c", &mark, sizeof(mark)); You must remove the trailing newline before attempting read with "%c" . A simple empty_stdin function will do,

eg

#include <stdio.h>

void empty_stdin ()
{
    int c = getchar();

    while (c != '\n' && c != EOF)
        c = getchar();
}
...
    printf("입력 : ");
    scanf_s("%d", &n1);
    empty_stdin();           /* remove all extraneous chars from stdin */
    printf("choose function : ");
    scanf_s("%c", &mark, sizeof(mark));

2. You fail to include break at the end of each switch case causing automatic fall-through to the next switch case, you need,

eg

    switch (mark)
    {
    case '+':
        result = Calc(n1, n2, Plus);
        break;
    case '-':
        result = Calc(n1, n2, Minus);
        break;
    }

Putting it altogether, you could do:

#include <stdio.h>

void empty_stdin ()
{
    int c = getchar();

    while (c != '\n' && c != EOF)
        c = getchar();
}

int Calc(int n1, int n2, int(*func) (int n1, int n2))
{
    return func(n1, n2);
}

int Plus(int n1, int n2)
{
    return n1 + n2;
}

int Minus(int n1, int n2)
{
    return n1 - n2;
}

int main(void)
{
    int n1, n2;
    int result = 0;
    char mark;

    printf("입력 : ");
    scanf_s("%d", &n1);
    empty_stdin();
    printf("choose function : ");
    scanf_s("%c", &mark, sizeof(mark));
    printf("입력 : ");
    scanf_s("%d", &n2);

    switch (mark)
    {
    case '+':
        result = Calc(n1, n2, Plus);
        break;
    case '-':
        result = Calc(n1, n2, Minus);
        break;
    }
    printf("%d\n", result);

    return 0;
}

( NOTE: you must also verify the return of each scanf to insure you have valid input -- that is left to you. Also note, you could ignore leading whitespace -- including '\\n' -- by including a space before your format specifier, eg " %c" , but that doesn't protect against a matching failure from your prior input -- ideally, you should empty_stdin() following each input with scanf to protect against that case and ready the input buffer for your next input.)

Example Use/Output

$ ./bin/fnpointer
입력 : 2
choose function : +
입력 : 3
5

Answer 2

您尚未初始化结果变量，您对n1，n2使用了scanf_s，但是对结果没有执行任何操作，因此编译器会抱怨