gesh.. those C-hacker.. Code like this is one of the reasons people don't talk to C-Programmer at cocktail parties! :-D
Anyway, while loops over its body as long as an expression is true. Now, it's body is just ";" so its empty.
But nobody said that the expression can't have any side effect. Here, the expression indeed assigns a value and increases a counter.
(ch=c[k++]) == ' '
When the compiler try to find out whether this is true or false, he sees, that this is actually the same like:
1 2
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char temp = ch=c[k++];
temp == ' ';
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Now the first statement is even composed of three things: Two assignments and one increment. Without the first assignment, you could write it like this*):
1 2 3
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ch=c[k++];
char temp = ch;
temp == ' ';
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And finally, "ch=c[k++]" are two statements and can be written as:
1 2 3 4
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ch=c[k];
k++;
char temp = ch;
temp == ' ';
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So it means: assign the value of c[k] to ch. Then increase k. If the value before the increase (stored in ch) is ' ', then still loop.
And that was only the first while loop. Now good luck figuring out how this fits into the outer loop.
Ciao, Imi.
*) That's not 100% right, but right enough for the case here ;-). The ignorable difference is, that actually = returns the second operand, not the first.