CODEWITHSUNDEEP
carraysstringinitialization
JGut
JGut

Reputation: 532

C - Array initialization using strlen making array wrong size

I'm experiencing an unexpected interaction with char array initialization.

When initializing a char[] with a size of strlen( char parameter[] ), the newly initialized char[] is too big. I am not sure what is causing this.

int main(void)
{
    char myString[] = { " " };
    foo( myString );
}    


int foo( char str[] )
{
    char testString[ strlen( str ) ];
    printf( "Length of testString: %lu\n", strlen( testString ) );
    return 0;
}

When I run foo, the output is 

Length of testString: 6

when I am expecting it to be 1.

Even stranger is when I add a print statement to foo before the initialization of testString, the output seems to magically fix itself:

int foo( char str[] )
{
    printf( "Length of str: %lu\n", strlen( str ) );
    char testString[ strlen( str ) ];
    printf( "Length of testString: %lu\n", strlen( testString ) );
    return 0;
}

foo now prints

Length of str: 1
Length of testString: 1

I have a feeling that it has something to do with the way char[] are passed into functions, or perhaps an unexpected behavior of strlen, but I really have no idea.

Any help is appreciated.

Upvotes: 4

Views: 1497

Answers (2)

user3629249
user3629249

Reputation: 16540

remember two things:

  1. strlen() yields the offset to the NUL char in the string and
  2. offsets into arrays start with 0, not 1

and now the code:

#include <string.h> // strlen(), strcpy()
#include <stdio.h>  // printf()

int main(void)
{
    char myString[] = { " " };  <<-- actual length 2 (a space + a NUL )
    foo( myString );
}    


int foo( char str[] )
{
    char testString[ strlen( str ) ]; <<-- strlen() yields 1, and testString not initialized
    printf( "Length of testString: %lu\n", strlen( testString ) ); <<-- strlen searchs until a NUL byte found which could be anywhere
    return 0;
}

Suggested corrections:

int foo( char str[] )
{
    char testString[ strlen( str )+1 ];
    strcpy( testString, str );
    printf( "Length of testString: %lu\n", strlen( testString ) );
    return 0;
}

Note: the output of the above/corrected code is 1, but the array testString[] is actually 2 bytes

another way to do it is:

int foo( char str[] )
{
    char testString[ strlen( str ) ] = {'\0'};
    printf( "Length of testString: %lu\n", strlen( testString ) );
    return 0;
}

then the output would be 0 and the actual length of testString[] is 1

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 310970

The array was not initialized

char testString[ strlen( str ) ];

Thus applying the function strlen to it results in undefined behavior.

printf( "Length of testString: %lu\n", strlen( testString ) );

Take into account that you may not initialize a variable length array like testString along with its declaration. You could write for example

char testString[ strlen( str ) ];
testString[0] = '\0';

It seems that what you mean is the sizeof operator. If so you can write

printf( "Length of testString: %zu\n", sizeof( testString ) );

Here is a demonstrative program

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

void foo( const char str[] )
{
    char testString[ strlen( str ) ];
    printf( "Length of testString: %zu\n", sizeof( testString ) );
}

int main(void) 
{
    char myString[] = { " " };
    foo( myString );

    return 0;
}

Its output is

Length of testString: 1

Upvotes: 5

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