Thread: Are these two pieces of code equivalent?

Is the first line of the following expression:

Code:

while((*(str + i) = *(str + j++)) != ‘\0’)
{
//code here
}

Equivalent to saying:

if (*(str + j) != ‘ ‘)
*(str + i) = *(str + j++);

??

It would really help me to know if it is so.

No. Even if you changed the '\0' to ' ' (or, I suppose, ' ' to '\0'), the first increments j and does the assignment statement no matter what, where the second checks first and then assigns.

What I don't understand is how this loop (the while one) is capable of eliminating spaces in a string with this code:

Code:

// Function to eliminate spaces from a string
void eatspaces(char* str)
{
int i = 0; // ‘Copy to’ index to string
int j = 0; // ‘Copy from’ index to string

while((*(str + i) = *(str + j++)) != ‘\0’) // Loop while character is not \0
if(*(str + i) != ‘ ‘) // Increment i as long as
i++; // character is not a space

return;
}

I know that the condition in this loop is supposed to copy the string by moving the character at position j to the character at position i and then increments j to the next character. If the character copied was ‘\0’, you have reached the end of the string and that's it; but I still don't get the ((*(str + i) = *(str + j++)) != ‘\0’) part and why it works. I don't how it actually copies the value in the whiel statement since, as you said it, the assignment occurs no matter what. I am confused.

Well, you have to actually read it. The character at str[j] is copied to str[i], and j is incremented (this is what gets us through the string). If it wasn't \0, go through the loop -- and the loop ONLY increments i if the character str[i] is not a space.

Yep, Both looks similar!

I've got it now tabstop. Feel like an idiot now. I read it but it got me confused for some reason. Thanks a lot!

Look at this,

Code:

 
 
 J =1;

 if ((i=j++)==1)
    printf("If i-%d,j-%d\n",i,j);  
 else
    printf("Else i-%d,j-%d\n",i,j);

It works in this way,

1. i=j
2. if (j == 1) results true
3. then increments the J to 2.

How different is this from the above eogoldseed's example....?

leogoldseed's example removes spaces from a string and yours does not. And J would only be 1 once...whereas in a string there might be several spaces... so even as a demonstration it's quite wrong.

Doing a similar thing with numbers might look like

Code:

#include <stddef.h>
#include <string.h>
int * removeif (int * to, int * from, size_t nelem, int value);

int main ()
{
    int foo[] = { 1, 2, 3, 4, 3, 2, 1, 1, 2 };
    int bar[10];
    
    memset(bar, 0, 10 * sizeof bar[0]);
    removeif(bar, foo, sizeof foo / sizeof foo[0], 1);

    return 0;
}

int * removeif (int * to, int * from, size_t nelem, int value)
{
    size_t i, j;

    for (i = 0, j = 0; i < nelem; i++) {
        if (value != from[i]) 
            to[j++] = from[i];
    }

    return to;
}