问题
I was expecting that in my following code:
#include<stdio.h>
int main(){
int i = 10;
int j = 10;
j = ++(i | i);
printf("%d %d\n", j, i);
j = ++(i & i);
printf("%d %d\n", j, i);
return 1;
}
expressions j = ++(i | i); and j = ++(i & i); will produce lvalue errors as below:
x.c: In function ‘main’:
x.c:6: error: lvalue required as increment operand
x.c:9: error: lvalue required as increment operand
But I surprised that above code compiled successfully, as below:
~$ gcc x.c -Wall
~$ ./a.out
11 11
12 12
Check the above code working correctly.
While other operators produce error (as I understand). Even bitwise operator XOR causes of an error j = ++(i ^ i); (check other operators produce an lvalue error at compilation time).
What is the reason? Is this is unspecified or undefined ? or bitwise OR AND operators are different?
compiler version:
gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5)
But I believe compiler version shouldn't reason for non-uniform behavior. If ^ not compiled then | and & also not. otherwise should work for all
Its not an error with this compiler in c99 mode: gcc x.c -Wall -std=c99.
回答1:
You are right that it should not compile, and on most compilers, it does not compile.(Please specify exactly which compiler/version is NOT giving you a compiler error)
I can only hypothesize that the compiler knows the identities that (i | i) == i and (i & i) == i and is using those identities to optimize away the expression, just leaving behind the variable i.
This is just a guess, but it makes a lot of sense to me.
回答2:
This is a bug that has been addressed in more recent GCC versions.
It's probably because the compiler optimizes i & i to i and i | i to i. This also explains why the xor operator didn't work; i ^ i would be optimized to 0, which is not a modifiable lvalue.
回答3:
C11 (n1570), § 6.5.3.1 Prefix increment and decrement operators
The operand of the prefix increment or decrement operator shall have atomic, qualified, or unqualified real or pointer type, and shall be a modifiable lvalue.C11 (n1570), § 6.3.2.1 Lvalues, arrays, and function designators
A modifiable lvalue is an lvalue that does not have array type, does not have an incomplete type, does not have a const- qualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a const- qualified type.C11 (n1570), § 6.3.2.1 Lvalues, arrays, and function designators
An lvalue is an expression (with an object type other thanvoid) that potentially designates an object.C11 (n1570), § 3. Terms, definitions, and symbols
Object: Region of data storage in the execution environment, the contents of which can represent values
As far as I know, potentially means "capable of being but not yet in existence". But (i | i) is not capable of referencing a region a data storage in the execution environment. Therefore it is not an lvalue. This seems to be a bug in an old gcc version, fixed since. Update your compiler!
回答4:
Just a follow-up to my question . I added elaborate answer so that one can find it helpful.
In my code expressions
j = ++(i | i);andj = ++(i & i);are not caused for lvalue error ?
Because of compiler optimization as @abelenky answered (i | i) == i and (i & i) == i. That is exactly CORRECT.
In my compiler (gcc version 4.4.5), any expression that includes single variable and result is unchanged; optimized into a single variable (something called not an expression).
for example:
j = i | i ==> j = i
j = i & i ==> j = i
j = i * 1 ==> j = i
j = i - i + i ==> j = i
==> means optimized to
To observe it I written a small C code and disassemble that with gcc -S.
C-Code: (read comments)
#include<stdio.h>
int main(){
int i = 10;
int j = 10;
j = i | i; //==> j = i
printf("%d %d", j, i);
j = i & i; //==> j = i
printf("%d %d", j, i);
j = i * 1; //==> j = i
printf("%d %d", j, i);
j = i - i + i; //==> j = i
printf("%d %d", j, i);
}
assembly output: (read comments)
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl $10, 28(%esp) // i
movl $10, 24(%esp) // j
movl 28(%esp), %eax //j = i
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl 28(%esp), %eax //j = i
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl 28(%esp), %eax //j = i
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl 28(%esp), %eax //j = i
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
In above assembly code all expressions converted into following code:
movl 28(%esp), %eax
movl %eax, 24(%esp)
that is equivalent to j = i in C code. Thus j = ++(i | i); and and j = ++(i & i); are optimized to j = ++i.
Notice: j = (i | i) is a statement where as expression (i | i) not a statement (nop) in C
Hence my code could successfully compiled.
Why
j = ++(i ^ i);orj = ++(i * i);,j = ++(i | k);produce lvalue error on my compiler?
Because either expression has the constant value or not modifiable lvalue (unoptimized expression).
we can observe using asm code
#include<stdio.h>
int main(){
int i = 10;
int j = 10;
j = i ^ i;
printf("%d %d\n", j, i);
j = i - i;
printf("%d %d\n", j, i);
j = i * i;
printf("%d %d\n", j, i);
j = i + i;
printf("%d %d\n", j, i);
return 1;
}
assembly code: (read comments)
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl $10, 28(%esp) // i
movl $10, 24(%esp) // j
movl $0, 24(%esp) // j = i ^ i;
// optimized expression i^i = 0
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl $0, 24(%esp) //j = i - i;
// optimized expression i - i = 0
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl 28(%esp), %eax //j = i * i;
imull 28(%esp), %eax
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl 28(%esp), %eax // j = i + i;
addl %eax, %eax
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 28(%esp), %edx
movl %edx, 8(%esp)
movl 24(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl $1, %eax
leave
Hence so this produce an lvalue error because operand is not a modifiable lvalue. And non-uniform behavior is due to compiler optimization in gcc-4.4.
Why new gcc compliers(or most of compilers) produce an lvalue error?
Because evaluation of expression ++(i | i) and ++(i & i) prohibits actual defination of increment(++) operator.
According to Dennis M. Ritchie's book "The C Programming Language" in section "2.8 Increment and Decrement Operators" page 44.
The increment and decrement operators can only be applied to variables; an expression like (i+j)++ is illegal. The operand must be a modifiable lvalue of arithmetic or pointer type.
I tested on new gcc compiler 4.47 here it produces error as I was expecting. I also tested on tcc compiler.
Any feedback/ comments on this would be great.
回答5:
I don't think at all it is an optimization error, because if it was, then there should not be any error in the first place. If ++(i | i) is optimized to ++(i), then there should not be any error, because (i) is an lvalue.
IMHO, I think that the compiler sees (i | i) as an expression output, that, obviously, outputs rvalue, but the increment operator ++ expects an lvalue to change it, thus the error.
来源：https://stackoverflow.com/questions/14860189/expressions-j-i-i-and-j-i-i-should-be-a-lvalue-error