Thread: array length

  1. #1
    Registered User
    Join Date
    Aug 2003

    array length

    I realized that strlen() only works on arrays of characters. I need to find a way to get the length of an array such as:
    int array[10];
    array[0] = 1;
    array[1] = 1;
    array[2] = 1;
    The answer would obviously be 3.
    I came up with this cheat that works in most cases:
    int length = 0;
    for (int x = 0; array[x]; ++x)
    return length;
    it works for the above example, but for cases like array[1] = 0; the for loop would terminate on array[1], causing the length to return as 1.

    What's a good way to find the length of any type of array?
    Thanks in advance.

    edit: Changing array[x]; to array[x] || array[x] == 0; won't work as there are a few ways my cheat can be disrupted.
    Last edited by Wick; 08-30-2003 at 02:19 PM.
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  2. #2
    Disturbed Boy gustavosserra's Avatar
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    Apr 2003
    As far as I know, you canīt do this. But to declare an array you must use a fixed size, either with static or dynamic allocation. I think that you could use the vector class if you want this kind of thing anyway!
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  3. #3
    Comment your source code! Lynux-Penguin's Avatar
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    Apr 2002
    int len = sizeof(myArray) / sizeof(myArray[0]);
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  4. #4
    Registered User
    Join Date
    May 2003
    This fails, however, when you pass the array to a function, which is why I never rely on that method.

    Wick, the best way is to not use arrays at all. Use a std::vector. It's as easy to use as an array, and you have the .size() method to tell you its size.


    #include <vector>
    #include <iostream>
    typedef std::vector<int> intArray;
    void printArray(intArray ia){
      for (int i = 0; i < (int) ia.size(); i++){
        std::cout << "[" << ia[i] << "]";
    int main(){
      intArray myArray(20); //creates an array of 20 elements
      for (int i = 0; i < (int) myArray.size(); i++)
        myArray[i] = 2* i;
    In 95% of cases, std::vector can be substituted freely for an array with minimal effort, and it's the better choice to use.

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