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length of array in function argument可能重复：函数参数中的数组长度
My array size is 5. For example:
我的数组大小为5.例如：
arrCustId[5]
How can I know how many Customer IDs are already present in my array? In short how to find length of array?
我怎么知道我的阵列中已存在多少个客户ID？总之如何找到数组的长度？
3 个解决方案
#1
26
If the array is statically allocated, use sizeof(array) / sizeof(array[0])
如果数组是静态分配的，请使用sizeof（array）/ sizeof（array [0]）
If it's dynamically allocated, though, unfortunately you're out of luck as this trick will always return sizeof(pointer_type)/sizeof(array[0]) (which will be 4 on a 32 bit system with char*s) You could either a) keep a #define (or const) constant, or b) keep a variable, however.
如果它是动态分配的，不幸的是，你运气不好，因为这个技巧总会返回sizeof（pointer_type）/ sizeof（array [0]）（在带有char * s的32位系统上将是4）你可以a）保持#define（或const）常数，或b）保持变量。
#2
12
Do you mean how long is the array itself, or how many customerids are in it?
你的意思是阵列本身有多长，或者有多少顾客？
Because the answer to the first question is easy: 5 (or if you don't want to hard-code it, Ben Stott's answer).
因为第一个问题的答案很简单：5（或者如果你不想硬编码，Ben Stott的回答）。
But the answer to the other question cannot be automatically determined. Presumably you have allocated an array of length 5, but will initially have 0 customer IDs in there, and will put them in one at a time, and your question is, "how many customer IDs have I put into the array?"
但是其他问题的答案无法自动确定。据推测，您已经分配了一个长度为5的数组，但最初会有0个客户ID，并且会将它们一次放入一个，您的问题是“我将多少客户ID放入数组？”
C can't tell you this. You will need to keep a separate variable, int numCustIds (for example). Every time you put a customer ID into the array, increment that variable. Then you can tell how many you have put in.
C不能告诉你这个。您需要保留一个单独的变量int numCustIds（例如）。每次将客户ID放入数组时，请递增该变量。然后你就可以知道你投入了多少。
#3
-8
I'm not sure that i know exactly what you mean.
我不确定我到底知道你的意思。
But to get the length of an initialized array,
但要获得初始化数组的长度，
doesn't strlen(string) work ??
不strlen（字符串）工作？
#1
26
If the array is statically allocated, use sizeof(array) / sizeof(array[0])
如果数组是静态分配的，请使用sizeof（array）/ sizeof（array [0]）
If it's dynamically allocated, though, unfortunately you're out of luck as this trick will always return sizeof(pointer_type)/sizeof(array[0]) (which will be 4 on a 32 bit system with char*s) You could either a) keep a #define (or const) constant, or b) keep a variable, however.
如果它是动态分配的，不幸的是，你运气不好，因为这个技巧总会返回sizeof（pointer_type）/ sizeof（array [0]）（在带有char * s的32位系统上将是4）你可以a）保持#define（或const）常数，或b）保持变量。
#2
12
Do you mean how long is the array itself, or how many customerids are in it?
你的意思是阵列本身有多长，或者有多少顾客？
Because the answer to the first question is easy: 5 (or if you don't want to hard-code it, Ben Stott's answer).
因为第一个问题的答案很简单：5（或者如果你不想硬编码，Ben Stott的回答）。
But the answer to the other question cannot be automatically determined. Presumably you have allocated an array of length 5, but will initially have 0 customer IDs in there, and will put them in one at a time, and your question is, "how many customer IDs have I put into the array?"
但是其他问题的答案无法自动确定。据推测，您已经分配了一个长度为5的数组，但最初会有0个客户ID，并且会将它们一次放入一个，您的问题是“我将多少客户ID放入数组？”
C can't tell you this. You will need to keep a separate variable, int numCustIds (for example). Every time you put a customer ID into the array, increment that variable. Then you can tell how many you have put in.
C不能告诉你这个。您需要保留一个单独的变量int numCustIds（例如）。每次将客户ID放入数组时，请递增该变量。然后你就可以知道你投入了多少。
#3
-8
I'm not sure that i know exactly what you mean.
我不确定我到底知道你的意思。
But to get the length of an initialized array,
但要获得初始化数组的长度，
doesn't strlen(string) work ??
不strlen（字符串）工作？