Int a = 5; a = a++ + ++a; a =? (2011)

Post by el_pollo_diablo » Thu, May 14, 2026, 9:20 PM UTC

IshKebab wrote: post #88

LPisGood wrote: post #57

Earlier quoted context omitted.

I don’t work a ton with C, but I wonder how C programmers keep track of what behavior is and is not defined. It seems like there are many possible edge cases.

They don't really. In fact there are many things that are technically UB but are so common that compilers can't really treat them as UB. E.g. type punning via unions.

Post by thaumasiotes » Thu, May 14, 2026, 9:24 PM UTC

Dylan16807 wrote: post #108

thaumasiotes wrote: post #105

Earlier quoted context omitted.

> It's Undefined Behavior. Susam's post doesn't make this clear. The quotes from K&R say that the modifications to the variable may take place in any order, but they don't directly say that doing this is Undefined Behavior, which would make it permissible to do anything, including e.g. interpreting the increments as decrements. The C99 standard is quoted saying this: >> Between the previous and next sequence point an…

Compilers are not able to prevent you from violating must/shall in the general case. So they're not held to that bar. Unless the standard says not to compile it, it's not a compiler bug. Also, imagine a situation where the line of code actually lists three different variables, but all three of them are passed in by address. It quickly becomes impossible for the compiler to know you violated the spec by reusing the sa…

Post by binaryturtle » Thu, May 14, 2026, 9:25 PM UTC

pavon wrote: post #95

binaryturtle wrote: post #81

Earlier quoted context omitted.

It's defined. And called "operator precedence", both post/pre-increment have a higher precedence than the single "+". At least according to this: https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Exp... I think the main confusion here comes from the fact that "a" is just a value, not a pointer, where it matters when the value/address which the pointer points at is accessed (before of after the increment of the…

Nope. Order of evaluation and operator precedence are completely unrelated. They should have been defined to be the same, but instead order of evaluation was left undefined. So if you write ++a + a++, operator precedence means this will be interpreted as (++a) + (a++), not say ++(a + a)++, but it is up to the compiler whether to execute ++a or a++ first, rather than executing them left to right.

    $ gcc -W -Wall test.c -o test
    test.c:8:7: warning: multiple unsequenced modifications to 'a' [-Wunsequenced]
            a = a++ + ++a;
                 ^    ~~
    1 warning generated.

Post by vcdk » Thu, May 14, 2026, 9:28 PM UTC

susam wrote: post #5

The code in the post seems very similar to the one in my own post from 2010: https://susam.net/sequence-points.html int a = 5; a += a++ + a++; I do remember that this particular code snippet (with a = 5, even) used to be popular as an interview question. I found such questions quite annoying because most interviewers who posed them seemed to believe that whatever output they saw with their compiler version was the co…

  $ /bin/cat x.c; gcc -w -o x x.c; ./x
  #include 
  
  int main()
  {
      int a = 5;
      a += a++ + a++;
      printf("a = %d\n", a);
  }
  a = 18

Post by chasil » Thu, May 14, 2026, 9:33 PM UTC

kencausey wrote: post #110

chasil wrote: post #69

Earlier quoted context omitted.

Awk also says it's 12. awk 'BEGIN{a=5; a = a++ + ++a; print a}' 12

https://www.gnu.org/software/gawk/manual/gawk.html#index-pre... "When side effects happen is implementation-defined. In other words, it is up to the particular version of awk."

  :/ $ awk 'BEGIN{a=5; a = a++ + ++a; print a}'
  12

  :/ $ which awk
  /system/bin/awk

  :/ $ awk --version
  awk version 20240728