C语言 如何在c中找到char数组的长度
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How to find the length of an char array in c
提问by stelios
I want to find the length of this :
我想找到这个的长度：
char *s[]={"s","a","b"};
it should count 4 with the /0 but the strlen or sizeof(s)/sizeof(char) gives me wrong answers.. How can i find it?
它应该用 /0 计算 4 但 strlen 或 sizeof(s)/sizeof(char) 给了我错误的答案..我怎么能找到它？
回答by Christian Rau
You are making an array of char*and not of char. That's why strlenwon't work. Use
您正在制作一个数组char*而不是char. 这就是为什么strlen行不通。用
sizeof(s) / sizeof(char*) //should give 3
If you want a single string use
如果你想要一个单一的字符串使用
char s[] = "sab";
回答by lhf
sizeof(s) / sizeof(s[0])works no matter what type scontains.
sizeof(s) / sizeof(s[0])无论s包含什么类型都有效。
回答by Naveen
What you have defined is not a string hence there is no NULL terminating character. Here you have declared pointers to 3 separate strings. BTW, you should declare your array as const char*.
您定义的不是字符串，因此没有 NULL 终止字符。在这里，您已经声明了指向 3 个单独字符串的指针。顺便说一句，您应该将数组声明为const char*.
回答by Zaur Nasibov
There is no direct way to determine the length of an array in C. Arrays in C are represented by a continuous block in a memory.
没有直接的方法可以确定 C 中数组的长度。 C 中的数组由内存中的连续块表示。
You must keep the length of the array as a separate value.
您必须将数组的长度保留为单独的值。
回答by atoMerz
strlen works if you terminate your array with null character. You cannot find number of elements in a char array unless you keep track of it. i.e store it in some variable like n. Every time you add member increment n and every time you remove decrement n
如果您使用空字符终止数组，则 strlen 有效。除非您跟踪它，否则您无法在 char 数组中找到元素的数量。即将它存储在一些像n这样的变量中。每次添加成员增加 n 和每次删除减少 n
回答by MByD
Why should it count 4? you have 3 pointers to char in this array, it should count 12 on most 32-bit platforms.
为什么要数4？在这个数组中有 3 个指向 char 的指针，在大多数 32 位平台上它应该是 12。
