This question already has answers here:
Using comma operator in c
(1 answer)
Not able to understand the reason for output
(3 answers)
Closed 8 years ago.
#include<stdio.h>
int main()
{
int x=10,y=12;
printf("%d",(x,y));
return 0;
}
The output of the program is 12. How?
The expression that you are evaluating is:
x,y
This expression uses the comma operator. The standard (6.5.17 Comma operator) says:
The left operand of a comma operator is evaluated as a void expression; there is a
sequence point between its evaluation and that of the right operand. Then the right
operand is evaluated; the result has its type and value.
So, in your code, x,y evaluates to y, which has a value of 12.
For a more expansive discussion, I refer you to cppreference.com. Although that discusses C++, the discussion for this operator is valid in the context of C. Particularly relevant to your question is this section:
The comma in various comma-separated lists, such as function argument lists (f(a, b, c)), initializer lists int a[] = {1,2,3}, or initialization statements (int i, j;) is not the comma operator. If the comma operator needs to be used in that context, it has to be parenthesized: f(a, (n++, n+b), c).
And that's exactly the situation in your question. If you had written:
printf("%d", x, y);
then there would have been no use of the comma operator, and you would have supplied one more argument to printf than format specifier.
You're by chance using the comma operator.
In the C and C++ programming languages, the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type).
That said,
printf("%d",(x,y));
is functionally equivalent to
printf("%d", y);
it is because first (x,y) is evaluated.
inside () the expression is x,y they are evaluated from left to right since the Associativity of Comma operator is left to right, so the last value of evaluating (x,y) is y.
read operator precedence and associativity rule and how expressions are evaluated under operator precedence to understand these type of expressions
Related
I came across this question and expected it to show up as a compile time error but to my surprise each statement separated by comma is executed and the final value is assigned to the variable.
int a,b=5;
a=(b++,++b,b*4,b-3);
printf("%d",a);
Output is 4
This output is exactly what should be the printed when each of those comma separated statements are executed separately. What I am not understanding is how and why does C allow this and how does compiler process this.
See What does the comma operator , do?
After understanding how the comma operator works, we can tell that this code is equivalent to:
int a,b=5;
b++;
++b;
b*4; // nonsense, the result isn't stored anywhere
a=b-3;
printf("%d",a);
5 + 1 + 1 - 3 = 4. The b*4 part does nothing and is just obfuscation.
comma acts both as a separator (in declaration of variables) and operator (evaluation of expressions).
In this case, comma acts as an operator. The comma operator introduces a sequence point between expressions.
The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value.
So your code is:
a = (b++, ++b, b*4, b-3);
Which is as if you had written this:
a = (5, 7, 28, 4);
So a is 4.
This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 6 years ago.
int main()
{
switch(1,2)
{
case 1:printf("1");break;
case 2:printf("2");break;
default: printf("error");break;
}
}
Is this valid in c?
I thought it shouldn't be , but when I compiled it , it shows no error and produces output 2.
Yes, this is valid, because in this case, the , is a comma operator.
Quoting C11, chapter §6.5.17, Comma operator, (emphasis mine)
The left operand of a comma operator is evaluated as a void expression; there is a
sequence point between its evaluation and that of the right operand. Then the right
operand is evaluated; the result has its type and value.
This (evaluates and) discards the left operand and uses the value of the right (side) one. So, the above statement is basically the same as
switch(2)
Just to elaborate, it does not use two values, as you may have expected something like, switching on either 1 or 2.
This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 7 years ago.
While completing a C programming test, I was given a question regarding the expected output from a function which seem to return two values. It was structured as follows:
int multi_return_args(void)
{
return (44,66);
}
The question caught me by surprise and inherently thought that if possible the first argument would be passed to the caller.
But after compiling it, the result is 66 instead. After a quick search I couldn't find anything about structuring a return statement like this so was wondering if some could help me.
Why does it behave like this and why?
The comma operator evaluates a series of expressions. The value of the comma group is the value of the last element in the list.
In the example you show the leading constant expression 44 has no effect, but if the expression had a side effect, it would occur. For example,
return printf( "we're done" ), 66;
In this case, the program would print "we're done" and then return 66.
In your code,
return (44,66);
is making (mis)use of the comma operator property. Here, it essentially discards the first (left side) operand of the , operator and returns the value of that of the second one (right operand).
To quote the C11 standard, chapter §6.5.17, Comma operator
The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value.
In this case, it is same as writing
return 66;
However, FWIW, the left hand side operand is evaluated as a void expression, meaning, if there is any side effect of that evaluation, that will take place as usual, but the result of the whole expression of the statement involving the comma operator will be having the type and the value of the evaluation of the right hand side operand.
This will return 66. There is nothing special about returning (44,66).
(44,66) is an expression that has the value 66 and therefore your function will return 66.
Read more about the comma operator.
Coma operator always returns the value of the rightmost operand when multiple comma operators are used inside an expression.
It will obviously return 66.
This question already has answers here:
C comma operator
(4 answers)
What does the comma operator , do?
(8 answers)
Closed 7 years ago.
I saw this in an exam and when I tried it out I was surprised. I tried it online and it works too. So I think it is the C language.
Why is that working? What is the use case for such an assignment syntax?
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char** argv) {
int i = (1,2,3,4,5);
printf("%d", i);
return 0;
}
These are not "multiple integers", but a comma operator. The whole parenthesised part is a single expression with each sub-expression (separated by commas) evaluated strictly from left to right. The results of all but the rightmost subexpression are ignored. The result of the whole expression is that of the last (rightmost) expression. Here it is the integer value 5.
Note that this operator is mostly used where only a single expression is allowed to add further side-effects. E.g. in a loop like:
int cnt = 0;
for ( const char *cp = "Hello" ; *cp != '\0' ; cp++, cnt++ ) ;
This counts the number of characters in a C-string, incrementing the pointer and cnt after each iteration. The results are ignored here.
So, this is in no way related to tuples or similar like in Python. There are actually no cases the usage of this operator is unavoidable and it should be used with caution — and some coding standards forbid its usage.
That's the comma operator at work. It evaluates the expression on its left-hand side, creates a sequence point, discards the value of the expression, then evaluates the expression on the right-hand side, and returns that as the value. When there are multiple expressions as in the example, then each is evaluated in turn and only the last is kept. With the sample, the compiler does the evaluation because every value is known at compile time. Note that the argument list to a function is not a use of the comma operator.
That isn't a valid use-case for the comma operator. What might be a more nearly valid use-case would be some operations with side-effects (such as function calls) that need to be sequenced and the final value assigned:
int i = (getchar(), getchar(), getchar());
This sets i to the third character in standard input, or EOF if there are not three characters left in the standard input to be read. Still not a realistic use-case, but better than assigning a list of constants.
In addition to the other answers, you need to watch for instances where a , is the comma operator as opposed to when it is a separator. For example, the following is invalid:
int i = 1,2,3,4,5;
In this case, the , is a separator between variable declarations. It declares i as an int and initializes it to 1, then it attempts to parse 2 as a variable name, which fails.
It works because you're using the "comma operator", which evaluates the subexpressions on the left and right, and has the value from the right-hand expression.
So in (1,2,3,4,5), 1 is evaluated and the result is discarded, then 2,3,4,5... in which (because of the next comma) 2 is evaluated and the result discarded, then 3,4,5... in which 3 is evaluated and discarded, then 4,5... in which 4 is evaluated and discarded, then 5 which becomes the result of the expression.
As for when it's useful, mainly as a shortcut when you need to evaluate several (sub)expressions for their side effects but aren't interested in their values (except maybe the last one). It's sometimes convenient in for loop expressions, such as when incrementing two variables:
for (i=0,j=1; j < len; i++,j++) {
..where it appears in both the initialization expression and the loop expression.
Why is that working?
Because its a valid C syntax. The comma in (1,2,3,4,5) are comma operator
C11: 6.5.17 Comma operator
Syntax
1 expression:
assignment-expression
expression , assignment-expression
Semantics
2 The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value.114)
What is the use case for such an assignment syntax?
See the example below
3 EXAMPLE As indicated by the syntax, the comma operator (as described in this subclause) cannot appear in contexts where a comma is used to separate items in a list (such as arguments to functions or lists of initializers). On the other hand, it can be used within a parenthesized expression or within the second
expression of a conditional operator in such contexts. In the function call
f(a, (t=3, t+2), c)
the function has three arguments, the second of which has the value 5.
void main()
int a,b,c;
c=(a,b)
This gives c=b while
c=a,b
gives c=a.
What is the reason for the above two?
In this line:
c=(a,b)
The parentheses mean, "evaluate the expression a,b first, then assign the value to c." In this case, b is assigned, because it's the right-hand-side expression of a,b. In C, comma expressions are evaluated left-to-right, with the overall value being that of the rightmost expression.
While in this line:
c=a,b
The assignment is evaluated as the entire left hand side first, which is c=a. This is because the equal = operator takes precedence over the comma , operator. Thus, b doesn't get assigned to c at all. It is equivalent to:
(c=a),b
In C, the comma operator evaluates the first operand, then discard it and then evaluates the right operand. So the outcome is the right operand. And it has the lowest precedence.
c = (a,b)
() has higher precedence than, so a,b evaluates first. The result is b. So c = b.
But when used c = a,b assignment = have higher precedence. So c = a evaluates first. Thus a is assigned to c.
Check this for further details.