将代码C ++转换为C. [英] Convert code C++ to C

问题描述

您好，我想将此行转换为C.此行基本上是为数组中的每个元素赋值。有谁能够帮我？非常感谢你

Hello, I want to convert this line to C. This line is basically assigning each element in the array a value. Can anybody help me? Thank you so much

char *arr = new char[10] {'A', 'K', 'U', 'E', 'B', 'Z', 'D', 'O', 'M', 'Q'};

我是什么尝试过：



我想这样做但是没有用。

What I have tried:

I wanted to do this but it is not working.

char *arr=(char *)malloc(11*sizeof(char)){'A', 'K', 'U', 'E', 'B', 'Z', 'D', 'O', 'M', 'Q'};

推荐答案

解决方案1是正确的，并且可以正常工作：但请记住该代码将生成一个局部变量：字符数组将在堆栈上，而不是堆：它只会在您从当前函数返回之前存在，并且不能以任何方式返回到调用函数而不会导致挂起引用 。如果在当前函数之外需要数组，则必须使用 malloc 并将该数组复制到它返回的基于堆的区域。

Solution 1 is correct, and will work: however do remember that that code will generate a local variable: the array of chars will be on the stack, not the heap: it will only exist until you return from the current function, and cannot be returned to the calling function in any way without causing a "hanging reference". If the array is needed outside the current function, you must use malloc and copy that array into the heap based area it returns.

char arr[] = {'A', 'K', 'U', 'E', 'B', 'Z', 'D', 'O', 'M', 'Q'};

这就是你所需要的do。

你甚至可以使用sizeof（arr）测试代码，你将得到10

That is all you need to do.
You can even test the code using sizeof(arr) and you will get 10

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