字符數組長度 function

Question

有誰知道如何編寫一個 function 來計算示例中給出的數組長度？

cout << length(argv[1]) << endl;

我試過這樣的東西，但它說那時不能有 function lentgh。

for(int i = 0; myStrChar[i] != '\0'; i++)
{
      count++;
}

EDIT1：這是代碼的 rest：

#include <iostream>
#include<cstring>
using namespace std;
int i;
int length()
{


for(int i = 0; myStrChar[i] != '\0'; i++)
{
      count++;
}
}
int main()
{

    cout << "Hello world!" << endl;
    cout << "length of first text: ";
cout << length(argv[1]) << endl;
char tab[255] = {"Ana has a cat"};

EDIT2：現在我做了這樣的事情，但它輸出 108 作為文本長度

char argv[]="ana has a cat";
int length(char l)
{
int cou= 0;

for(int i = 0; i!=0; i++)
{

     cou++;
     return cou;
}
}

Answer 1

嗯，怎么了？ 這是一個有效的代碼 - 只需將其包裝到 function 中：

size_t length(const char* myStrChar)
{
    size_t count = 0;
    for(int i = 0; myStrChar[i] != '\0'; i++) { 
        count++;
    }
    return count;
}

你可以做的更短一點：

size_t length(const char* myStrChar)
{
    size_t count = 0;
    for (; myStrChar[count] != 0; count++)
        /* Do nothing */;
    return count;
}

UPD：這是來自問題的代碼和我對作者的評論：

// You're passing just a single character to this function - not array
// of char's - not a string.
// Pass char* or better const char*.
int length(char l)
{
int cou= 0;

// Ok, this loop does nothing - it's going to stop at the
// first condition check, because you assign 0 to i
// and immediately check if it is not zero - this check will
// return false and the loop will never start.
// Check the current symbol in string if it is not zero
for(int i = 0; i!=0; i++)
{

     cou++;

     // Even if you'll fix mistakes above, your loop will stop
     // at the first iteration, because of this return.
     // You must put it right after the loop.
     return cou;
}

// Your function actually returns nothing, because no iterations
// of the loop above is performed - so no return statement reached
// at all - it is undefined behaviour :(
// Put return here out of the loop;
}