Consider the following C++ code:<\/p>\n
\r\nvoid foo(int a, int b, int c) \r\n{ \r\n std::cout << a << ' ' << b << ' ' << c << std::endl; \r\n} \r\n\r\nint main(int argc, char* argv[]) \r\n{ \r\n int value = 1; \r\n foo(value++, value++, value++); \r\n\r\n return 0; \r\n}<\/pre>\nThat looks pretty straight-forward: there is a function that takes three integer arguments and prints them to the console. In main, it is called by incrementing a variable three times. You would expect that the output was 1 2 3. But surprise: it is 3 2 1 in a debug build and 1 1 1 in a release build (Visual Studio 2005 and Visual Studio 2008). Why? Because it writes multiple times to the same memory location between two sequence points, which is undefined behavior (and I know somebody that always stresses that undefined behavior is spelled W-R-O-N-G).<\/p>\n
A sequence point is a point in the execution of the program where all side effects from previous evaluation have been performed and no side effects from subsequent evaluation have been performed. Between two consecutive sequence points an object’s value can be modified only once by an expression. The function call operator is such a sequence point, but the order of arguments evaluations is not specified (unlike Java or C# where is always performed from left to right). Thus, modifying a variable multiple times in calling a function introduces undefined behavior.<\/p>\n
Using the \/RTCu (Run-time error checks) compiler option in the debug build produces different machine code and order of evaluation.<\/p>\n
At machine code level, the code for main in the debug build looks like this:<\/p>\n
\r\n; 34 : \r\n\r\n\tmov\tDWORD PTR _value$[ebp], 1 \r\n\r\n; 35 : \r\n\r\n\tmov\teax, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv65[ebp], eax \r\n\tmov\tecx, DWORD PTR _value$[ebp] \r\n\tadd\tecx, 1 \r\n\r\n\tmov\tDWORD PTR _value$[ebp], ecx \r\n\tmov\tedx, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv68[ebp], edx \r\n\tmov\teax, DWORD PTR _value$[ebp] \r\n\tadd\teax, 1 \r\n\r\n\tmov\tDWORD PTR _value$[ebp], eax \r\n\tmov\tecx, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv71[ebp], ecx \r\n\tmov\tedx, DWORD PTR _value$[ebp] \r\n\tadd\tedx, 1 \r\n\r\n\tmov\tDWORD PTR _value$[ebp], edx \r\n\tmov\teax, DWORD PTR tv65[ebp] \r\n\tpush\teax \r\n\tmov\tecx, DWORD PTR tv68[ebp] \r\n\tpush\tecx \r\n\tmov\tedx, DWORD PTR tv71[ebp] \r\n\tpush\tedx \r\n\tcall\t?foo@@YAXHHH@Z\t\t\t\t; foo \r\n\tadd\tesp, 12\t\t\t\t\t; 0000000cH<\/pre>\nand in the release build (or without \/RTCu):<\/p>\n
\r\n; 34 : \r\n\r\n\tmov\tDWORD PTR _value$[ebp], 1 \r\n\r\n; 35 : \r\n\r\n\tmov\teax, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv65[ebp], eax \r\n\r\n\tmov\tecx, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv68[ebp], ecx \r\n\r\n\tmov\tedx, DWORD PTR _value$[ebp] \r\n\tmov\tDWORD PTR tv71[ebp], edx \r\n\r\n\tmov\teax, DWORD PTR tv65[ebp] \r\n\tpush\teax \r\n\tmov\tecx, DWORD PTR tv68[ebp] \r\n\tpush\tecx \r\n\tmov\tedx, DWORD PTR tv71[ebp] \r\n\tpush\tedx \r\n\tcall\t?foo@@YAXHHH@Z\t\t\t\t; foo \r\n\tadd\tesp, 12\t\t\t\t\t; 0000000cH \r\n\r\n\tmov\teax, DWORD PTR _value$[ebp] \r\n\tadd\teax, 1 \r\n\tmov\tDWORD PTR _value$[ebp], eax \r\n\r\n\tmov\tecx, DWORD PTR _value$[ebp] \r\n\tadd\tecx, 1 \r\n\tmov\tDWORD PTR _value$[ebp], ecx \r\n\r\n\tmov\tedx, DWORD PTR _value$[ebp] \r\n\tadd\tedx, 1 \r\n\tmov\tDWORD PTR _value$[ebp], edx<\/pre>\nIf you know a little bit assembly language you can see that in the first case value is incremented after each evaluation of foo’s arguments, and in the second case that happens only after the call to foo. After the call, in both cases, value becomes 4.<\/p>\n
To achieve the intended behavior you should write:<\/p>\n
\r\nint main(int argc, char* argv[]) \r\n{ \r\n int value = 1; \r\n foo(value, value+1, value+2);\r\n value += 3;\r\n\r\n return 0; \r\n}<\/pre>\nIt should be quite obvious, that the same behavior is encountered if the call to foo was replaced with:<\/p>\n
\r\nstd::cout << value++ << ' ' << value++ << ' ' << value++ << std::endl;<\/pre>\nFor more information about sequence points I suggest reading:
\nhttp:\/\/en.wikipedia.org\/wiki\/Sequence_point<\/a>
\nhttp:\/\/c-faq.com\/expr\/seqpoints.html<\/a>
\nhttp:\/\/msdn2.microsoft.com\/en-us\/library\/d45c7a5d(VS.80).aspx<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"Consider the following C++ code: void foo(int a, int b, int c) { std::cout << a << ‘ ‘ << b << ‘ ‘ << c << std::endl; } int main(int argc, char* argv[]) { int value = 1; foo(value++, value++, value++); return 0; } That looks pretty straight-forward: there is a function that takes … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","advgb_blocks_editor_width":"","advgb_blocks_columns_visual_guide":"","_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[7],"tags":[],"class_list":["post-83","post","type-post","status-publish","format-standard","hentry","category-c"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t