expression-statement question

[Chad]

Give the following...

#include <stdio.h>
#include <stdlib.h>

struct point{
int x;
int y;
};

int main(void)
{

struct point *points;

if ((points = malloc(sizeof(struct point))) == NULL) {
fprintf(stderr, "unable to allocate memory\n");
exit(1);
}

free(points);

exit(0);
}

Doesn't the if statement only expressions and not expression-
statements? Just curious, because (points = malloc(sizeof(struct
point)) is an expression-statement. How does this get converted to an
expression so that the if statement can use it?

Chad

 

[Ben Pfaff]

if ((points = malloc(sizeof(struct point))) == NULL) { [...]
Doesn't the if statement only expressions and not expression-
statements?
Yes.

Just curious, because (points = malloc(sizeof(struct point)) is
an expression-statement.

No, it's just an expression (if you add the missing ")" at the
end).

 

[Keith Thompson]

Give the following... [snip]
if ((points = malloc(sizeof(struct point))) == NULL) { [...]
}

Doesn't the if statement only expressions and not expression-
statements? Just curious, because (points = malloc(sizeof(struct
point)) is an expression-statement. How does this get converted to an
expression so that the if statement can use it?

(You're missing a verb in that first sentence, but I can from context.)

No, ``points = ...'' is not an expression statement, it's just an
expression. An assignment operator is just another operator (that
happens to have a side effect), and an assignment is an expression.
It yields the value of the target object after the assignment.

An expression *statement* is a statement consisting of an expression
followed by a semicolon. (The grammar also treats a null statement,
consisting of just a semicolon, as an expression statement; I'm
not sure why.)

It happens that assignment expressions most commonly appear in
expression statements:
x = 42;
but that's really just a special case.

(In some other languages, an assignment is a statement but not an
expression.)

 

[Chad]

Give the following... [snip]
  if ((points = malloc(sizeof(struct point))) == NULL) { [...]
  }

Doesn't the if statement only expressions and not expression-
statements? Just curious, because (points = malloc(sizeof(struct
point)) is an expression-statement. How does this get converted to an
expression so that the if statement can use it?

(You're missing a verb in that first sentence, but I can from context.)

No, ``points = ...'' is not an expression statement, it's just an
expression.  An assignment operator is just another operator (that
happens to have a side effect), and an assignment is an expression.
It yields the value of the target object after the assignment.

An expression *statement* is a statement consisting of an expression
followed by a semicolon.  (The grammar also treats a null statement,
consisting of just a semicolon, as an expression statement; I'm
not sure why.)

It happens that assignment expressions most commonly appear in
expression statements:
    x = 42;
but that's really just a special case.

(In some other languages, an assignment is a statement but not an
expression.)

--
Keith Thompson (The_Other_Keith) (e-mail address removed)  <http://www.ghoti.net/~kst>
Nokia
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

The original question was meant to be worded as 'Doesn't the if
statement only allow expressions and not expression-statements?'.

 

[Seebs]

The original question was meant to be worded as 'Doesn't the if
statement only allow expressions and not expression-statements?'.

Okay. The answer is, yes, but you're confused. Assignment has nothing
to do with whether or not something is a statement. A semicolon does.

You can't write:
if (3
because the if statement takes an expression, not an expression-statement.

-s