strlen 为二维数组返回错误值

Question

Why is this:

char animals[][30] = {{"cow "}, {"dog "}, {"frog "}, {"gecko "}, {"cat"}};

returning 4 instead of 5 when I do this:

printf("%d\n", strlen(animals));

From what I believe, there are 5 different arrays of character arrays so i'm not sure why it is saying there are 4.

Answer 1

This declaration

char animals[][30] = {{"cow "}, {"dog "}, {"frog "}, {"gecko "}, {"cat"}};

does not declare a one-dimensiobal array that contains a string.

The type of the argument of the function strlen shall be char * . But you are using an argument of the type char ( * )[30] .

What you need is the following

printf( "%zu\n", sizeof( animals ) / sizeof( *animals ) );

i'm not sure why it is saying there are 4.

Because in this call of printf that uses an invalid conversion specifier %d instead of %zu

printf("%d\n", strlen(animals));

the value of the argument animals in the call

strlen(animals)

is considered as the address of the first element of the array that indeed has the length equal to 4 .

That is from the point of view of equality of addresses this call

strlen(animals)

is equivalent to the call

strlen(animals[0])

because the value of animals used in the first call is equal to the value of the address of its first element animals[0] . Though in the first call of strlen the argument has an invalid type.

The compiler should issue a message that you are using an incorrect argument in the call of strlen .

Answer 2

strlen() is for calculating the length of strings (null-terminated sequence of characters in C), not for calculating the number of elements of array.

To obtain the number of elements in an array, you can divide the size of array by the size of an element of the array.

printf("%zu\n", sizeof(animals) / sizeof(*animals));

Also note that you should use %zu , not %d , to print size_t ( strlen() and sizeof return size_t )

Answer 3

There are multiple issues with the call printf("%d\n", strlen(animals));

• the printf format should be %zu if strlen has been properly declared by including <string.h> at the point of the call. This is unlikely to pose a problem but should be fixed.
• strlen() takes a C String and animal is not a C string, it is a 2D array of char . You should instead pass strlen(animal[0]) . Again this mistake is unlikely to pose a problem because the address of the first string in the 2D array is the same as that of the array itself. To compute the number of elements in the 2D array, use sizeof(animal) / sizeof(*animal) .
• the call should print 4 (the string "cow " has a trailing space) and a newline, and the return value will be 2 as the number of characters output by printf .

Try this modified version:

#include <stdio.h>
#include <string.h>

int main() {
    char animals[][30] = {{"cow "}, {"dog "}, {"frog "}, {"gecko "}, {"cat"}};

    printf("number of strings: %zu\n", sizeof(animals) / sizeof(*animals));
    for (size_t i = 0; i < sizeof(animals) / sizeof(*animals); i++) {
        printf("%zu: \"%s\", %zu chars\n", i, animals[i], strlen(animals[i]));
    }
    return 0;
}

Answer 4

Basic answer to your question

"returning 4 instead of 5 when I do this"

when you do this

printf("%d\n", strlen(animals));

animals is a pointer to the first of the strings, animals is stored in memory as "cow \0........dog \0..." where..... are bytes until the next array entry (usually random memory stuff if not initialized). When you printf strlen of animals it returns strlen of "cow ".

You are mixing types.. c allows this because they are pointers really, this is why compiler works but it is wrong.

to get 5 as result you will need to do this

printf("%zu",sizeof(animals) /sizeof(animals[0]));