为什么编译器没有显示*标记被用作标记[I]的错误？ [英] Why doesn't the compiler show an error for *marks being used as marks[I]?

问题描述

#include< stdio.h>

int main（）

{

int * marks;

printf（输入四个标记：）;

for（int i = 0; i< 4; i ++）

scanf（％d，& marks [i]）;

printf（你输入的标记是：）;

for（int i = 0; i< 4; i ++）

printf（％d，marks [i]）;

返回0;

}



我尝试了什么：



程序没有给出任何输出也没有显示任何错误

#include <stdio.h>
int main()
{
int *marks;
printf("Enter four marks:");
for(int i = 0;i<4;i++)
scanf("%d",&marks[i]);
printf("Marks you entered are:");
for(int i=0;i<4;i++)
printf("%d ",marks[i]);
return 0;
}

What I have tried:

The program did not give any output neither did it show any error

推荐答案

你宣布

int * marks;



你没有为它分配任何内存。 &NBSP;你有点幸运，你的应用程序不只是崩溃。



类似

marks =（int *）malloc（4 * sizeof （int））;

You declared
int *marks;

You didn't allocate any memory for it.   You're somewhat fortunate your application didn't just crash.

Something like
marks = (int *)malloc(4*sizeof(int));

引用：

为什么编译器没有显示*标记被用作标记[I]的错误？

Why doesn't the compiler show an error for *marks being used as marks[I]?

因为语言定义说的是数组的名称是指向第一个元素的指针。所以标记同时是一个指针和一个数组;使用方式可以通过声明来实现。

Because the language definition says that the name of an array is a pointer to the first element. So marks is both a pointer and an array t the same time; the usage is available either way you declare it.

int arr[10];
int i1 = arr[0];
int i2 = *arr;

全部有效，原样：

Is all valid, as is:

int *arr = (int*) malloc(40);
int i1 = arr[0];
int i2 = *arr;

这篇关于为什么编译器没有显示*标记被用作标记[I]的错误？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！