查找字符数组的长度 [英] Finding length of char array

问题描述

我有一些代码如下:

int i=0;
char a[7]={0x00,0xdc,0x01,0x04};
int len=0;
len = sizeof(a);
printf("The Length is : %d", len);

这里我想在c中找到数组a的长度?这怎么办?

Here I want to find the length of the array a in c? How can this be done?

推荐答案

按照约定，C 字符串是以空字符结尾的".这意味着末尾有一个额外的字节，其值为 0 (0x00).任何对字符串执行某些操作的函数(如 printf)都会在它找到 null 时认为字符串结束.这也意味着如果你的字符串不是空终止的，它会一直运行直到找到一个空字符，这会产生一些有趣的结果！

By convention C strings are 'null-terminated'. That means that there's an extra byte at the end with the value of zero (0x00). Any function that does something with a string (like printf) will consider a string to end when it finds null. This also means that if your string is not null terminated, it will keep going until it finds a null character, which can produce some interesting results!

由于数组中的第一项是 0x00，因此它将被视为长度为零(无字符).

As the first item in your array is 0x00, it will be considered to be length zero (no characters).

如果您将字符串定义为:

If you defined your string to be:

char a[7]={0xdc,0x01,0x04,0x00};

例如空终止

然后就可以使用strlen来测量存储在数组.

then you can use strlen to measure the length of the string stored in the array.

sizeof 测量类型的大小.这不是你想要的.还要记住，数组中的字符串可能比数组的大小还短.

sizeof measures the size of a type. It is not what you want. Also remember that the string in an array may be shorter than the size of the array.

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