c - C:将数组传递给函数并在接收者函数中进行迭代

Question

我想将数组传递给C中的函数并对其进行迭代。我有此代码：

#include <stdio.h>

int funct(int * a);

int main(int argc, char ** argv){
    int a[5] = {0};
    int b[5] = {1, 1};

    printf("Size of cache: %d\n", sizeof(a));
    printf("Array values:\n");
    printf("Numb of elments in a[]: %d\n", (sizeof(a) / sizeof(a[0])));

    for(int i = 0; i < (sizeof(a) / sizeof(a[0])); i++){
        printf("for loop\n");
        printf("%d\n", a[i]);
    }
    printf("\n");

    printf("Size of cache: %d\n", sizeof(b));
    printf("Array values:\n");
    printf("Numb of elments in a[]: %d\n", (sizeof(b) / sizeof(b[0])));

    for(int i = 0; i < (sizeof(b) / sizeof(b[0])); i++){
        printf("for loop\n");
        printf("%d\n", b[i]);
    }
    printf("\n");

    funct(a);
    funct(b);

    return 0;
}

int funct(int * a){

    printf("Size of cache: %d\n", sizeof(a));
    printf("Numb of elements in a[]: %d\n", (sizeof(a) / sizeof(a[0])));
    printf("Array values:\n");

    for(int i = 0; i < (sizeof(a) / sizeof(a[0])); i++){
        printf("sizeof(a): %d\n",sizeof(a));
        printf("sizeof(a[0]): %d\n",sizeof(a[0]));
        printf("for loop\n");
        printf("%d\n", a[i]);
    }
    printf("\n");

    return 0;
}

结果是：

Size of cache: 20
Array values:
Numb of elments in a[]: 5
for loop
0
for loop
0
for loop
0
for loop
0
for loop
0

Size of cache: 20
Array values:
Numb of elments in a[]: 5
for loop
1
for loop
1
for loop
0
for loop
0
for loop
0

Size of cache: 4
Numb of elements in a[]: 1
Array values:
sizeof(a): 4
sizeof(a[0]): 4
for loop
0

Size of cache: 4
Numb of elements in a[]: 1
Array values:
sizeof(a): 4
sizeof(a[0]): 4
for loop
1

请解释为什么我不能遍历函数内部的数组-我在做什么错（i）以及如何正确地做到这一点（ii）。谢谢

Answer 1

因为sizeof运算符为您提供指针的大小，而不是数组中的元素数，所以您应该这样编写函数

int funct(int * a, int count){

    printf("Size of cache: %d\n", sizeof(a));
    printf("Numb of elements in a[]: %d\n", (sizeof(a) / sizeof(a[0])));
    printf("Array values:\n");

    for(int i = 0; i < count; i++){
        printf("sizeof(a): %d\n",sizeof(a));
        printf("sizeof(a[0]): %d\n",sizeof(a[0]));
        printf("for loop\n");
        printf("%d\n", a[i]);
    }
    printf("\n");

    return 0;
}

然后在 main()函数调用中

funct(a, sizeof(a) / sizeof(a[0]));
funct(b, sizeof(b) / sizeof(b[0]));

您无法获得指针指向的元素数，因此唯一的方法是将其作为函数参数与数组一起传递。

另请注意，您没有正确初始化数组，并且在尝试打印数组的元素时这将是未定义的行为。