打印char数组C ++ [英] Printing char arrays c++

问题描述

我在C ++中玩弄c字符串，发现一些我不终止char数组时不了解的行为.

char strA[2] = {'a','\0'};
char strB[1] = {'b'};
cout << strA << strB;

我希望它会打印ab，但是会打印aba.如果我改为在strA之前声明strB，它将按预期工作.有人可以解释这是怎么回事吗?

解决方案

这是未定义的行为，您很幸运地替换了这两个数组的声明对您有用.让我们看看您的代码中发生了什么:

char strA[2] = {'a','\0'};

创建一个可以像字符串一样对待的数组-它以 null终止.

char strB[1] = {'b'};

创建一个数组，该数组不能被视为字符串，因为它缺少空终止符'\0'.

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std::cout << strA << strB;

第一部分是<< strA，可以正常工作.因为strA被当作const char*对待，所以它会打印a，作为std::ostream& operator <<的参数提供的参数将被用于打印每个字符，直到遇到空终止符为止.. >

那会发生什么?然后，执行<< strB(实际上，与将该行分为两个单独的std::cout <<调用相比，实际上有点不同，并且更复杂，但是在这里没关系).它也被视为const char*，，预计将以提到的'\0' 结尾，但不是...

这会导致什么?您很幸运，内存中'\0'之前(还是-随机)只有一个字符，这停止了(可能接近无限)打印过程.

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为什么，如果我改为在strA之前声明strB，它会按预期工作?

那是因为您很幸运，编译器决定在strB之后声明您的strA，因此，在打印strB时，它会打印包含它的所有内容并打印strA，最后具有空终止符.这不能保证.避免使用char[]表示和打印字符串.请改用std::string，它会为您处理空终止符.

I was playing around with c strings in c++ and found some behavior I don't understand when I don't terminate a char array.

char strA[2] = {'a','\0'};
char strB[1] = {'b'};
cout << strA << strB;

I would expect this to print ab, but instead it prints aba. If I instead declare strB before strA, it works as expected. Could someone explain what's going on here?

解决方案

This is undefined behaviour and you simply are lucky that replacing the declaration of these 2 arrays works for you. Let's see what is happening in your code:

char strA[2] = {'a','\0'};

Creates an array that can be treated like a string - it is null terminated.

char strB[1] = {'b'};

Creates an array that cannot be treated like a string, because it lacks the null terminating character '\0'.

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std::cout << strA << strB;

The first part, being << strA, works fine. It prints a since strA is treated as a const char*, which provided as an argument for std::ostream& operator << will be used to print every character untill the null terminating character is encountered.

What happens then? Then, the << strB is being executed (actually what happens here is a little different and more complicated than simply dividing this line into two, separate std::cout << calls, but it does not matter here). It is also treated as a const char*, which is expected to be ended with mentioned '\0', however it is not...

What does that lead to? You are lucky enough that there randomly is only 1 character before (again - random) '\0' in memory, which stops the (possibly near-infinite) printing process.

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Why, if I instead declare strB before strA, it works as expected?

That is because you were lucky enough that the compiler decided to declare your strA just after the strB, thus, when printing the strB, it prints everything that it consists + prints strA, which ends with null terminating character. This is not guaranteed. Avoid using char[] to represent and print strings. Use std::string instead, which takes care of the null terminating character for you.

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