CODEWITHSUNDEEP
c
darsie
darsie

Reputation: 147

operand evaluation order + assignment side effect

I want the following code to print 11, but prints 12, except in the last case, where it prints 10.

x=5; x1=x+(x=6); printf("%d\n",x1);  
x=5; x1=(x=x)+(x=6); printf("%d\n",x1);  
x=5; x1=(x+0)+(x=6); printf("%d\n",x1);  
x=5; x1=(x=5)+(x=6); printf("%d\n",x1);  

x=5; x1=(x=6)+x; printf("%d\n",x1);  
x=5; x1=(x=6)+(x=x); printf("%d\n",x1);  
x=5; x1=(x=6)+(x+0); printf("%d\n",x1);  
x=5; x1=(x=6)+(x=5); printf("%d\n",x1);

gcc says in every case: 'warning: operation on ‘x’ may be undefined'.

That's mean.

Bernhard

PS: There's no question, sorry. Thanks for your answers. :)
PPS: Actual code is: 

while ( data-(data=read(adr)&(1<<6)) ) i++;

I'm waiting for bit 6 at adr to stop toggling.

Upvotes: 1

Views: 671

Answers (3)

caf
caf

Reputation: 239051

You can use the little-used comma operator, along with another variable, in order to write the loop you wanted:

while ( lastdata = data, lastdata != (data = read(adr) & (1<<6)) ) i++;

Upvotes: 1

Matthew Flaschen
Matthew Flaschen

Reputation: 284836

There's a reason for the warning... The evaluation order between sequence points is unspecified.

Upvotes: 2

anon
anon

Reputation:

The results are undefined, no further explanation necessary. But to explain two possible ways the compiler could treat your code:

int x = 1;
int n = (x=3) + x;

The compiler can evaluate (x=3) first in which case the assignment to n has the value 6. Or it can evaluate x first, in which case the assignment to n has the value 4.

Upvotes: 2

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