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

问题描述

我有一些code如下：

I have some code as follows

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

在这里，我想找到在C数组的长度？
我该怎么办呢？

Here I want to find the length of the array a in c? How can I do it?

推荐答案

按照惯例C字符串是空值终止。这意味着，有一个在与零值结束一个额外的字节（ 0×00 ）。那做了一个字符串（如的printf ）的任何函数会考虑一个字符串结束时，发现空。这也意味着，如果你的字符串不是空终止，它会继续下去，直到找到一个空字符，它可以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!

由于你的数组中的第一项为0x00，它会被认为是零长度（没有字符）。

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

如果您定义您的字符串是：

If you defined your string to be:

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

例如。空值终止

那么你可以使用 的strlen 来衡量长度存储在数组中的字符串

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

的sizeof 衡量一个类型的大小。这是不是你想要的。还记得在数组中的字符串可能比数组的长度短。

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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