[Ana24]

int  a = 5;
int b = a++;
int c = ++a;
printf ("%d %d %d\n", a, b, c);
return 0

I don't understand how the out put is supposed to be 7 5 7 when the first is a and a=5
Thanks

[macosxnerd101]

The a++ and ++a operations increment the value stored in the variable. The a++ operation evaluates after the value of a (5) is assigned to b. The ++a operation is executed before the value of a (now 7) is assigned to c.

Also, remember to end your return statement with a semi-colon.

[jimblumberg]

The variable 'a' starts at five then you increment 'a' twice. After the increments 'a' will be seven.

Jim

[Ana24]

macosxnerd101, on 07 February 2012 - 08:30 AM, said:

The a++ and ++a operations increment the value stored in the variable. The a++ operation evaluates after the value of a (5) is assigned to b. The ++a operation is executed before the value of a (now 7) is assigned to c.

Also, remember to end your return statement with a semi-colon.

Thanks, but shouldn't it be 5 5 7 ? I don't understand how the first changes...

[jimblumberg]

Here is the same program without the increment operators:

/*
int  a = 5;
int b = a++;
int c = ++a;
printf ("%d %d %d\n", a, b, c);
return 0;
*/
int a = 5;
int b = a;
a = a + 1; // to replace a++;
a = a + 1; // This will replace the ++a
c = a;
printf ("%d %d %d\n", a, b, c);

Jim

[macosxnerd101]

The variable a is 7 at the point of the printf() statement, the variable b takes on 5 before a becomes 6, then a becomes 7 and that value is assigned to c.