为什么我不需要scanf的&符? (在C中) [英] Why don't I need ampersands with the scanf? (In C)

问题描述

void getnums(int *a, int *b);

int main()   
{  
    int a;
    int b;
    int c;
    getnums(&a,&b);
    c = a + b;
    printf("a + b = %d\n", c);
    return 0;  
}

void getnums(int *a, int *b)
{ 
    printf("a:? ");
    scanf("%d", a);
    printf("b:? ");
    scanf("%d", b);
}

为什么在scanfs中的a和b之前不需要&符? (该代码当前有效.)

解决方案

因为scanf将指针作为其参数(以便它知道要修改的变量)，并且a和b已经是指针. /p>

void getnums(int *a, int *b);

int main()   
{  
    int a;
    int b;
    int c;
    getnums(&a,&b);
    c = a + b;
    printf("a + b = %d\n", c);
    return 0;  
}

void getnums(int *a, int *b)
{ 
    printf("a:? ");
    scanf("%d", a);
    printf("b:? ");
    scanf("%d", b);
}

Why don't I need ampersands before the a and b in the scanfs? (The code currently works.)

解决方案

Because scanf takes pointers as its arguments (so that it knows what variable to modify), and a and b are already pointers.

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