Getting wrong answer when counting array in c

Question

#include <stdio.h>
#include <stdlib.h>


int countArrayChars(char *strArray[]){
    int i=0;
    while (strArray[i] != '\0'){
        i++;
    }
    printf("%d\n", i);
    return i;
}

int main(int argc, const char * argv[]) {
    char *dog[] = {"dog"};
    countArrayChars(dog);

For some reason, it prints "5".

Shouldn't it print 3? I even tried to put \0 after the "g".

or char dog[] = {"dog"}; ..int countArrayChars(char strArray[]){ — BLUEPIXY
– BLUEPIXY, Commented Oct 14, 2015 at 0:58
Like BLUEPIXY said, strArray is an array of strings so strArray[0] is a pointer to the first string in the array and strArray[1] doesn't exist. What you are iterating is the bytes in the address of the pointer to the string. — Jerry Jeremiah
– Jerry Jeremiah, Commented Oct 14, 2015 at 0:59
Can you give a quick/simple explanation of why having two [ ] [ ] fixed it? Sorry, sort of a beginner. — Boris R
– Boris R, Commented Oct 14, 2015 at 1:13

ashiquzzaman33 · Accepted Answer · 2015-10-14 01:56:33Z

1

You declare array of string and initialize it with dog.

char *dog[] = {"dog"};

Actually it represented as

dog[0] = "Dog";     //In your case only element index with 0.
...............
...............
dog[n] = "tiger";   //If there Have n+1 element

Hence your array size is 1. Which hold constant string dog. To access it you should use dog[0].

So without less modification you can use your code as:

int countArrayChars(char *strArray[])
{
    int i=0;
    while (strArray[0][i] != '\0')
    {
        i++;
    }
    printf("%d\n", i);
    return i;
}

int main(int argc, const char * argv[])
{
    char *dog[] = {"dog"};
    countArrayChars(dog);
}

Or if you want to declare a string use

char *dog = "dog";

or

char dog[] = "dog";

edited Oct 14, 2015 at 1:56

answered Oct 14, 2015 at 1:31

ashiquzzaman33

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Comments

BobRun · Accepted Answer · 2015-10-14 01:24:42Z

0

Please try this

 #include <stdio.h>
 #include <stdlib.h>


int countArrayChars(char *strArray){
    int i=0;
    while (strArray[i] != '\0'){
        i++;
    }
    printf("%d\n", i);
    return i;
}

int main(int argc, const char * argv[]) {
    char *dog[] = "dog";
    countArrayChars(dog);
    }

answered Oct 14, 2015 at 1:24

BobRun

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2 Comments

BobRun Over a year ago

Sorry to disagree, it compiles and runs and gives 3.

BobRun Over a year ago

Could you try a C compiler ? That's what I am using, not C++

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