Pass-by-value weirdness

[sillyfofilly]

I'm still trying to get my head completely around pointers, and so I've been writing a few programs to see what does and doesn't work. This code is to find anagrams. It takes in a one word argument, and checks it against a list of words (words.txt, a sample of which I have attched. The real file is too big to attach here).

If you run the following code

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

int isAnagram(char word1[], char word2[]);
void removeFirst(char letter, char* string[]);
void stripWord(char* junk[]);

static const int yes = -1;
static const int no = 0;

int main(int argc, char* argv[]) {
	if ( argc != 2 ) {
		printf("Please enter exactly one argument\n");
		return -1;
	} else if ( strlen(argv[1]) > 20 ) {
		printf("Please enter an argument that is at most 20 characters\n");
		return -1;
	}

	char word[20];

	strcpy(word, argv[1]);
	
	static const char filename[] = "words.txt";
	FILE *file = fopen ( filename, "r" );

	char line [ 128 ];
	char tmp[128];
	while ( fgets ( line, sizeof line, file ) != NULL ) {
	
		int i;

		for ( i = 0; i < 256; i++ )
		{
			if ( line[i] == '\n' )
			{
				line[i] = '\0';
				break;
			}
		}
	
	
		//sprintf(tmp,"%s",word);
		printf("%s\n",word);
		if ( isAnagram(line,word) == yes ) printf("%s\n", line);
		printf("%s\n",word);
	}
	
	fclose ( file );

	return 0;
}

int isAnagram(char word1[], char word2[]) {

	if ( strlen(word1) != strlen(word2) ) return no;
	if ( strcmp(word1,word2) == 0 ) return no;
	
	int i;
	for (i=0; i<strlen(word1);i++) {
		if ( strchr(word2,word1[i]) == NULL ) return no;
		else {
			removeFirst(word1[i],&word2);
		}
	}
	
	return yes;
}

void removeFirst(char letter, char* string[]) {
	//remove the first occurrence of 'letter' in 'word'
	
	char temp[20]="";
	int i,index;
	
	index = strchr(*string,letter)-*string;
	
	for (i=0; i<strlen(*string);i++) {
		if ( i != index ) sprintf(temp,"%s%c",temp,(*string)[i]);
	}
	strcpy(*string,temp);
}

The problem I'm having is at these lines

//sprintf(tmp,"%s",word);
printf("%s\n",word);
if ( isAnagram(line,word) == yes ) printf("%s\n", line);
printf("%s\n",word);

When it runs like that, the output will show you the value of the variable 'word', before and after the execution of the isAnagram function. This value appears to change as the program runs.

However, if you uncomment the sprintf line, and pass the value tmp instead of word, like so

sprintf(tmp,"%s",word);
printf("%s\n",word);
if ( isAnagram(line,tmp) == yes ) printf("%s\n", line);
printf("%s\n",word);

the value of word doesn't change at all, and it works fine.

So, my question is, Why is 'word' being changed by isAnagram when I didn't pass it's address to the function?

Thank you

[MK27]

So, my question is, Why is 'word' being changed by isAnagram when I didn't pass it's address to the function?

Thank you

Perhaps we understand "didn't" differently, because I would say that you did so:

if ( isAnagram(line,word) == yes ) printf("%s\n", line);

You need to make a local copy in the function if you don't want changes to affect the original variable -- or send a copy to the function, as w/ "tmp".

You can't pass an array by value.

[sillyfofilly]

Ok, so passing it like this

if ( isAnagram(line,word) == yes ) printf("%s\n", line);

allows for the value to be changed? I thought that automatically created a local copy of the variable, and to specifically allow it to be changed you had to pass the address like this

if ( isAnagram(line,&word) == yes ) printf("%s\n", line);

[vart]

word pointer is not changed, buffer pointed by the word pointer could be changed

[matsp]

All arrays in C are passed "by reference" (or put another way, as the address of the first element). &word would be a pointer to pointer to char.

It also means that your removefirst doesn't need to be char *string[].
--
Mats

[MK27]

Ok, so passing it like this

if ( isAnagram(line,word) == yes ) printf("%s\n", line);

allows for the value to be changed? I thought that automatically created a local copy of the variable, and to specifically allow it to be changed you had to pass the address like this

if ( isAnagram(line,&word) == yes ) printf("%s\n", line);

No. The only reason you got away with the second one is because of this:

void removeFirst(char letter, char* string[])

which is very wrong, and the only reason that works is because you are working on the first element. I think if you try and do something to the second element here, you will notice it's the forth letter. That's because char* string[] is an array of pointers, not an array of characters. If you try this, for example

if ( isAnagram(line,&word) == yes ) printf("%s\n", line);

You will get an error.
So to make it clear again: You can't pass an array by value in C, and a string is an array of characters.

[Snafuist]

So to make it clear again: You can't pass an array by value in C

And to make it confusing again: C always uses call-by-value.

If you pass a pointer-to-array to a function, you are calling the function "by value". But it's the pointer's value that is passed, not the value of the array. Consider the following code:

#include <stdio.h>

void foo(int *bar)
{
        bar = 3;
}

int main()
{
        int arr[1024];
        int *p;

        p = arr;

        printf("%p\n", p);
        foo(p);
        printf("%p\n", p);

        return 0;
}

This will print the same value of p twice, because foo() only modifies its local copy of p, namely bar. The array values at arr will remain unchanged. On the other hand, if foo() would instead be "*bar = 3;", this would actually change the first value of arr in main().

Maybe it helps to think of pointers as notes with a number written on them. If you pass such a note to a function, you're actually making an exact copy of the note (call-by-value). The function is free to do anything with it, including overwriting the address on the note. The calling function won't be able to tell (because it has its own note), and of course this won't change the stuff at the specified address. On the other hand, going to the specified address and changing values there will of course have the same effect in the calling function.

Hope this helps.

Greets,
Philip

[King Mir]

&word would be a pointer to pointer to char.

In main it wouldn't be. It would be a pointer to an array of chars. So you wouldn't be able to call removeFirst from main.

In isAnagram, word is a char pointer, so it works.

[MK27]

Maybe it helps to think of pointers as notes with a number written on them. If you pass such a note to a function, you're actually making an exact copy of the note (call-by-value). The function is free to do anything with it, including overwriting the address on the note. The calling function won't be able to tell (because it has its own note), and of course this won't change the stuff at the specified address. On the other hand, going to the specified address and changing values there will of course have the same effect in the calling function.

What that means is if you reassign the pointer in the called function:

void somefunc(char *string) {
         char new[]="this";
         string=new;
}

This will not have any affect outside, because the local copy of the "pointer value" (an address) has changed, so operations on string no longer affect the variable at the address string originally referred to. Also, string won't contain the array it originally did. But if you don't reassign the pointer, then changes to string are global.

By the way, these mean the same thing:

void somefunc(char *string) {
void somefunc(char string[]) {

which is different from

void somefunc(char* string[]) {

Also, you may want to try Snafuist's code as this demonstrates the use of printf's handy %p, which will probably help you to get more of a grip on the pointer concept.