C++ overloaded operator argument evaluation order

Asked Mar 27 '15 at 17:16

Active Mar 27 '15 at 17:22

Viewed 49 times

The readByte and readInt are inline methods. The problem is that the readInt call is actually executed first, so i get wrong OPCODE, cause it is already poped by thereadInt call. Is it normal? I think they should be performed in order as they appear in source code.

log << "OPCODE: " << std::hex << (unsigned int)bytes.readByte() << ", DIFF: " << bytes.readInt() << std::endl;

edited Mar 27 '15 at 17:22

user207421

305,947
44
307
483

asked Mar 27 '15 at 17:16

Krab

6,526
6
41
78

2

This has nothing to do with operator precedence. It's the order of evaluation of operands. – chris Mar 27 '15 at 17:17
There is only one `readByte()` call here. – user207421 Mar 27 '15 at 17:18
The `<<` operator is evaluated left-to-right. That is just how the world turns. – riodoro1 Mar 27 '15 at 17:19
2

@chris It is about the order of evaluation of *operands.* – user207421 Mar 27 '15 at 17:19
@EJP, My bad, I was thinking about functions (which are close enough that it's a terminology issue here). – chris Mar 27 '15 at 17:20
1

@riodoro Not so. See the duplicate. – user207421 Mar 27 '15 at 17:25
@riodoro1, Note that I use two different compilers here: http://coliru.stacked-crooked.com/a/4ac6e925bab1ded6 – chris Mar 27 '15 at 17:38

C++ overloaded operator argument evaluation order

0 Answers0