Thread: C comma operator problem

Well what did you expect?

You type in "y\n"

The second time around the loop, c == '\n' and the test fails, exit loop.

This is pretty poor use of the comma operator BTW.

As in
while ( scanf("%c\n",&c) == 1 && c == 'y' )

Someone knows if there's an equivalent to the comma in others programming language like Java or Basic ?

Java has only a restricted version of the comma operator, which can be used in the initialization and increment sections of for loops. It cannot be applied in while loops, do-while loops, or the conditional section of for loops -- boolean logic like || or && must be applied there. The designers of Java thought that allowing the comma operator anywhere else allowed programmers to create code that is too hard to read.

Personally I use the comma operator occasionally, usually when I repeat the same simple operation on multiple variables:

Code:

if(lower_right()) x ++, y ++;

Code:

char c;
while(scanf("%c\n",&c)==1 && c=='y')
printf("Hello");

//or

char c;
while(scanf("%c\n",&c), c=='y')
printf("Hello");

I mean either of the above snippets do not work in the first loop. The i/o is like this
y
y
Hello

You see my point.


and I have one more thing to discuss.

I read somewhere the precise syntax of a for loop is
for(condition 1;condition 2;expression)
code

and i happened to test it by writing code like

Code:

int i=0;
for(i==0;i<5;i++)
printf("Hello");

and it worked.

but the strange part is I believe an assignment statement return the r value for eg. i=1 returns 1 and according to this i=0 should return 0 and so our traditional

Code:

for(i=0;i<5;i++)
code

shouldn't work. Well how does it work then??

> I mean either of the above snippets do not work in the first loop.
That's because \n doesn't mean what you think it means.
It does NOT mean read the next character (and check it is newline), it means absorb any arbitrary sequence of whitespace characters. So "\n" and "\n\n\n \t\t\t\t\n" would both be matched entirely by a single \n in the conversion. What is more, it can't return until it sees a character which isn't whitespace, which explains why nothing happens until you type in the second 'y'.

while(scanf("%c ",&c)==1 && c=='y')
has exactly the same behaviour.

Learn to read all input with fgets(), then examine the buffer in memory. Learning all the intricacies of just scanf is about as hard as learning the rest of C.

> I read somewhere the precise syntax of a for loop is
> for(condition 1;condition 2;expression)
Well the first one is wrong, it should be initialisation.

> for(i==0;i<5;i++)
Which would be the same as
for( ; i<5 ; i++ )
You compare i with 0, but since there is nothing to assign or test the result, it is simply thrown away.
But since you've already assigned i = 0 outside the loop, all you really have is
for ( i = 0 ; i < 5 ; i++ )