Length of array of strings

Question

How can we determine the length of an array of strings when we don't know the length?

For example in this piece of code:

#include <stdio.h>

int main() {
    int n;
    char names[3][10] = { “Alex”, “Phillip”, “Collins” };

    for (n = 0; n < 3; n++)    
        printf(“%s \n”, names[n]);
}

n < 3 is assuming you know the length of the array but how can you determine it's length when we don't know?

I have tried a few alternatives such as:

int arraySize() {
    size_t size, i = 0;
    int count = 0;
    char names[3][10] = { “Alex”, “Phillip”, “Collins” };

    size = sizeof(names) / sizeof(char*);

    for (i = 0; i < size; i++) {
        printf("%d - %s\n", count + 1, *(names + i));
        count++;
    }

    printf("\n");
    printf("The number of strings found are %d\n", count);
    return 0;
}

or

for (n = 0; n < sizeof(names); n++)

but they all error out.

Any help is appreciated.

Given what you know about sizeof, what do you think sizeof names produces? What does sizeof names[0] produce? What does sizeof names[0][0] produce? Do those give you any idea? — Eric Postpischil
– Eric Postpischil, Commented Nov 14, 2021 at 16:12

chqrlie · Accepted Answer · 2021-11-14 16:26:02Z

3

I am not sure what you mean by they all err out.

There is a syntax issue in the code posted as you use guillemet characters instead of double quotes: “Alex” should be "Alex". This could be a side effect of your cut/paste method to post the code, but nevertheless a potential issue.

Your approach using size = sizeof(names) / sizeof(char*); is right but the type is incorrect: names[0] is not a char *, it is an array of 10 characters. You should use size = sizeof(names) / sizeof(names[0]); which works for all arrays, regardless of the type.

Here is a modified version where the length of the array is determined by the compiler:

#include <stdio.h>

int main() {
    char names[][10] = { "Alex", "Phillip", "Collins" };
    int i, length = sizeof(names) / sizeof(names[0]);

    for (i = 0; i < length; i++)    
        printf("%d: %s\n", i + 1, names[i]);

    return 0;
}

Notes:

you could use size_t instead of int for array length and index variables, but it is only necessary for very large arrays and the printf conversion specifier would be %zu for a value of type size_t.
it is less confusing to use length for the length of an array and reserve size for sizes in bytes obtained from sizeof().

edited Nov 14, 2021 at 16:26

answered Nov 14, 2021 at 16:21

chqrlie

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3 Comments

RandomDeveloper Over a year ago

thanks! And yes it was a copy and paste issue, my code has the right syntax. Nevertheless, this sorted my issue, thanks very much

RandomDeveloper Over a year ago

I have tried to use my function as an "external function" to main but getting loads of errors now:

int arraySize(char names[]) {     size_t size, i = 0;     int count = 0;      size = sizeof(names) / sizeof(names[0]);      for (i = 0; i < size; i++) {         count++;     }     printf("%d", count);     return count; }  int main() {     char names[3][10] = {"Alex", "Philip", "John"};    arraySize(names); }

any ideas?

chqrlie Over a year ago

The names argument is not an actual array, the function receives a pointer to the first element, from which the length of the original array cannot be determined. sizeof(a)/sizeof(a[0]) can only be used with an array, not a pointer or a function argument.

sourav sharma · Accepted Answer · 2021-11-14 16:18:29Z

-1

first every string is ended with a special character '\0' means

{'A','L','E','X','\0'}

even you haven't put '\0' there but compiler put it there for it convenient

int lenght(char *str){
int count = 0 ;
for(int i=0;str[i]!='\0';i++){
    count++ ;
}
return count ;

}

use this function to count your string length like :

int main(){
char names[3][10] = {"Alex", "Phillip", "Collins"};
printf(“%d \n”,lenght(names[1] );
return 0 ;
}

answered Nov 14, 2021 at 16:18

sourav sharma

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