为什么会出现在我的计划无限循环？ [英] Why is there an infinite loop in my program?

问题描述

int main(void)
{
    int i;
    int array[5];

    for (i = 0; i <= 20; i++)
    array[i] = 0;

    return 0;
}

为什么上面的code停留在一个无限循环？

Why is the above code stuck in an infinite loop?

推荐答案

下面是happenning在给定的code。

Here is what happenning in the given code.

#include<stdio.h>
#include<string.h>
int main(void)
{
    int i;
    int array[5];

    for (i = 0; i <= 20; i++)
    {
        printf("%p %p \n",&i,&array[i]);
        printf("the value of i is %d \n",i);
        sleep(1);
        array[i] = 0;
        printf("i may be modified here lets see what i is %d \n", i);
    }

    return 0;
}

在我的堆栈内存给我的地址位置为

in my stack memory I got the address locations as

I 存储在位置0xbfd1048c地址

i is stored at location 0xbfd1048c address

和阵列存储在位置0xbfd10478地址

and array is stored at location 0xbfd10478 address

当你在一个时间点递增 I 值为每个循环数组的地址[I] 相当于地址 I （它只是指针引用）

As you are incrementing i value for each loop at one point of time the address of array[i] is equivalent to address of i (its just pointer dereferencing)

您在数组[我] 存储了那么什么是什么，但 I 的实例地址，所以你是在写 I 的实例值，以0作为你提到的数组[我] = 0 相当于到 I = 0 这样的条件 I＆LT; = 20 总是成功

So what you are storing at array[i] is nothing but the i's instance address so you are over writing the i's instance value to 0 as you have mentioned array[i] = 0 which is equivalent to i=0 so the condition i<=20 always succeeds.

现在为什么内存以这样一种方式分配的大问题。

Now the BIG question why does the memory allocated in such a way.

这是在运行时间和资源的可用性内核决定。

It is decided at run time and on the availability of the resources to the kernel.

所以这就是为什么我们与数组的限制居住。

So that's why we have to dwell with in the limits of the array.

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