WHEN YOU TYPE

Code:

void func(char *s1, char *s2, char *s3){
 while(*s1 != '\0' ) s1++;
 while(*s2 != '\0' ) s2++;
 for(; *s1=*s2=*s3; s1++, s2++, s3++);
}

main() {
 char s1[10] = "ABC", s2[10]="DEF";
 char s3[10] = "GHI";
 func(s1,s2,s3);
 printf("%s, %s %s", s1, s2, s3);
 }

this code gives ABCGHI, DEFGHI GHI as the result....
how come there is a command assignment in the middle of for function
and why is the result like that? I would be very much appreciated if
you tell me step by step

come on experts!!! I'm desperate...

This

Code:

for(initialization; condition; increment) { /* ... */ }

is the same as

Code:

initialization;
while(condition) {
    /* ... */
    increment;
}

or

Code:

initialization;
value = condition;
while(value) {
    /* ... */
    increment;
    value = condition;
}

It follows that this

Code:

for(; *s1=*s2=*s3; s1++, s2++, s3++);

is the same as

Code:

while(*s1=*s2=*s3) {
    s1++, s2++, s3++;
}

or

Code:

value = *s1=*s2=*s3;
while(value) {
    s1 ++;
    s2 ++;
    s3 ++;
    value = *s1=*s2=*s3;
}

Oh, and this

Code:

while(*s1 != '\0' ) s1++;

is the same as

Code:

value = (*s1 != '\0');
while(value) {
    s1 ++;
    value = (*s1 != '\0');
}

Does that help answer your question?

[edit] Hint: loops like this

Code:

while(*s1 != '\0' ) s1++;

just increment s1 until s1 is pointing at the NULL terminator. Thus, after the first two lines of func(), s1 and s2 have both been incremented to point at their NULL terminators. [/edit]

You post again after 4 minutes? Don't be lame. This isn't a drive-through fast food service, and even those might take longer than 4 minutes.

You posted a function that must be related to a programming course or something else designed to teach you, because it would be overly difficult to read otherwise.

As with the answer, I compiled it myself and got this:

Code:

ABCGHI, DEFGHI GHI

Inside func, basically s1 and s2 will point to the end of the strings they respectively point to.

When the for loop occurs, this is what you have:

Code:

         s1
          |
          |
          v
A, B, C, \0


         s2
          |
          |
          v
D, E, F, \0

s3
|
|
v
G, H, I, \0

From there on, the condition of the for loop performs a copy. This is dangerous to do. In this program, it's guarenteed to work (I think), but the function assumes it's being passed large enough buffers, which can result in problems (similar to gets()).

opps.... the result is ABCGHI, DEFGHI GHI... I'm sorry mistake

>this code gives ABCGHI, DEFGHI DEF as the result....
I suggest a new compiler.

you guys are awesome... thank you so much!!!!

Very strange, but my output is
"ABCGHI, DEFGHI GHI"

I think - you miswrited it...

also main should be
int main(void)

Now I've got another problem...
at the end of for statement, pointer s1, s2, s3are all pointing to the '\0' character of each
string.
when you come back to main and print each string why does it print the whole thing? is it supposed to???
and oh, another one.
When does the for statement officially end??
how does it get "false" value to end the statement???

There's no answer to that level of impatience.

Oh wait, there is.
http://catb.org/~esr/faqs/smart-questions.html#urgent

I answered these new questions in the other topic....

>When does the for statement officially end??
At the semicolon.
>how does it get "false" value to end the statement???
A false is a zero, so when the value of *s3 becomes zero (the string terminator is found), then *s2 will be assigned to 0, as will *s1, so the expression evaluates to 0, or false.

I'm not sure I can do a good job of explaining question 1, so I'll defer.

for ends when condition is false

This *s1=*s2=*s3 is false when the return value is 0, the return value is the value assigned to *s1. So when the null-character from the string s3 is copied to s2 ans s1 - for loop ends

As to the first questing - parameters in C are passed by value, so original pointers are not changed, func function changes only the local copies of the pointers s1,s2,s3