[Spiffster]

Yet another question from a beginner reading a book.

What is wrong with this code:

char *p[][2] = {
…string table stuff…
};

int main()
{
int i;
char str[80];

printf("Enter apple name: ");
fgets(apple, 80, stdin);

/*search array */
for(i=0; *p[i][0]; i++){
if(!strcmp(apple, *p[i][0]))
printf("%s and %s.", str, *p[i][1]);
}
}

I copied it just like it says in my book, and when it gets the the for loop it quits with a bus error. It doesn't seem to look for a bad memory address. Please help me! (again ).

[lgerbarg]

Originally posted by Spiffster:
Yet another question from a beginner reading a book.

What is wrong with this code:

Code:
char *p[][2] = {
	…string table stuff…
	};
	
int main()
{
	int i;
	char str[80];
	
	printf("Enter apple name: ");
	fgets(apple, 80, stdin);
	
	/*search array */
	for(i=0; *p[i][0]; i++){
              if(!strcmp(apple, *p[i][0]))
			printf("%s and %s.", str,  *p[i][1]);
        }
}

I copied it just like it says in my book, and when it gets the the for loop it quits with a bus error. It doesn't seem to look for a bad memory address. Please help me! (again ).

The simple answer is the code is wrong, and any book that uses it as an example is not a good book to be learning from.

The longer answer is that its using arrays in a way that is basically pointer arithmatic. And its wrong. You could fix those bugs by changing the for loop to:

Code:
	/*search array */
	for(i=0; p[i][0]; i++){
              if(!strcmp(str, p[i][0]))
			printf("%s and %s.", str,  p[i][1]);
        }
}

And you also have to make sure that the last entry in the array is a NULL. Even that will probably not get you what you want, since the fgets will have a "\n" at the end of i, and chances are the strings in your array do not.

So if you want a fully working example you get:

Code:
#include <stdio.h>

char *p[][2] = {
        "apple\n",
        "test",
        NULL,
        NULL
};

int main()
{
        int i;
        char str[80];

        printf("Enter apple name: ");
        fgets(str, 80, stdin);

        /*search array */
        for(i=0; p[i][0]; i++){
                if(!strcmp(str, p[i][0]))
                printf("%s and %s.", str, p[i][1]);
        }
}

[tofueater:~] local% ./a.out 
Enter apple name: apple
apple
 and test.[tofueater:~] local%

Of course you probably want to strip the return from the end of string, but the way this code is arranged the only consistent way would be:

Code:
#include <stdio.h>

char *p[][2] = {
        "apple",
        "test",
        NULL,
        NULL
};

int main()
{
        int i;
        char str[80];

        printf("Enter apple name: ");
        fgets(str, 80, stdin);

        str[strlen(str)-1] = 0;

        /*search array */
        for(i=0; p[i][0]; i++){
                if(!strcmp(str, p[i][0]))
                printf("%s and %s.", str, p[i][1]);
        }
}

[tofueater:~] local% 
[tofueater:~] local% ./a.out
Enter apple name: apple
apple and test.[tofueater:~] local%

Again, this is not good fix, its just consistent with the exisiting code. I am somewhat curious what the book is, and what the chapter is about. Unless this from a chapter on pointer aliasing or something of the kin I would say the book is just totally rotten, and even then its really questionable code.

Louis

[Spiffster]

Actually, it was my fault. I tried to rewrite the example. It originally used array indexes, and i tried to rewrite it using pointer artihmatic. I guess i fscked up pretty baddly 8^).

[lgerbarg]

Originally posted by Spiffster:
Actually, it was my fault. I tried to rewrite the example. It originally used array indexes, and i tried to rewrite it using pointer artihmatic. I guess i fscked up pretty baddly 8^).

Its not bad for someones first time trying it, I thought it was actually a published example from a book. I would not have been quite so harsh had I realized that.

The thing you want to remember is that in C arrays are just syntactic sugar for pointer math. So in your above code:

a[i] is equivalent to (char *)(*(a + i) ) and
a[i][j] is equivalent to (char *)(*(a + i + j))

In fact that causes one of those fun oddities of C:

a[i][j] == a[j][i] == i[a][j] == j[i][a] .....

Louis