{"id":4712,"date":"2022-08-24T01:14:54","date_gmt":"2022-08-23T19:44:54","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-does-px-x-always-result-in-p-for-pointer-p-and-integer-x-in-gcc-linux-x86-64-c\/"},"modified":"2022-08-24T01:14:54","modified_gmt":"2022-08-23T19:44:54","slug":"solved-does-px-x-always-result-in-p-for-pointer-p-and-integer-x-in-gcc-linux-x86-64-c","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-does-px-x-always-result-in-p-for-pointer-p-and-integer-x-in-gcc-linux-x86-64-c\/","title":{"rendered":"[Solved] Does (p+x)-x always result in p for pointer p and integer x in gcc linux x86-64 C++"},"content":{"rendered":"

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Yes, for gcc5.x and later specifically, that specific expression is optimized very early to just p<\/code>, even with optimization disabled, regardless of any possible runtime UB.<\/p>\n

This happens even with a static array and compile-time constant size. gcc -fsanitize=undefined<\/code> doesn’t insert any instrumentation to look for it either. Also no warnings at -Wall -Wextra -Wpedantic<\/code><\/p>\n

int *add(int *p, long long x) {\n    return (p+x) - x;\n}\n\nint *visible_UB(void) {\n    static int arr[100];\n    return (arr+200) - 200;\n}\n<\/code><\/pre>\n
Using gcc -dump-tree-original<\/code> to dump its internal representation of program logic before any optimization passes shows that this optimization happened even before that in gcc5.x and newer<\/strong>.  (And happens even at -O0<\/code>).<\/p>\n
;; Function int* add(int*, long long int) (null)\n;; enabled by -tree-original\n\nreturn <retval> = p;\n\n\n;; Function int* visible_UB() (null)\n;; enabled by -tree-original\n{\n  static int arr[100];\n\n    static int arr[100];\n  return <retval> = (int *) &arr;\n}\n<\/code><\/pre>\n
That’s from the Godbolt compiler explorer with gcc8.3<\/a> with -O0<\/code>.<\/p>\n
The x86-64 asm output is just:<\/p>\n
; g++8.3 -O0 \nadd(int*, long long):\n    mov     QWORD PTR [rsp-8], rdi\n    mov     QWORD PTR [rsp-16], rsi    # spill args\n    mov     rax, QWORD PTR [rsp-8]     # reload only the pointer\n    ret\nvisible_UB():\n    mov     eax, OFFSET FLAT:_ZZ10visible_UBvE3arr\n    ret\n<\/code><\/pre>\n
-O3<\/code> output is of course just mov  rax, rdi<\/code><\/strong><\/p>\n
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gcc4.9 and earlier only do this optimization in a later pass, and not at -O0<\/code><\/strong>: the tree dump still includes the subtract, and the x86-64 asm is<\/p>\n
# g++4.9.4 -O0\nadd(int*, long long):\n    mov     QWORD PTR [rsp-8], rdi\n    mov     QWORD PTR [rsp-16], rsi\n    mov     rax, QWORD PTR [rsp-16]\n    lea     rdx, [0+rax*4]            # RDX = x*4 = x*sizeof(int)\n    mov     rax, QWORD PTR [rsp-16]\n    sal     rax, 2\n    neg     rax                       # RAX = -(x*4)\n    add     rdx, rax                  # RDX = x*4 + (-(x*4)) = 0\n    mov     rax, QWORD PTR [rsp-8]\n    add     rax, rdx                  # p += x + (-x)\n    ret\n\nvisible_UB():       # but constants still optimize away at -O0\n    mov     eax, OFFSET FLAT:_ZZ10visible_UBvE3arr\n    ret\n<\/code><\/pre>\n
This does line up with the -fdump-tree-original<\/code> output:<\/p>\n
return <retval> = p + ((sizetype) ((long unsigned int) x * 4) + -(sizetype) ((long unsigned int) x * 4));\n<\/code><\/pre>\n
If x*4<\/code> overflows, you’ll still get the right answer.  In practice I can’t think of a way to write a function that would lead to the UB causing an observable change in behaviour.<\/p>\n
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As part of a larger function, a compiler would be allowed to infer some range info, like that p[x]<\/code> is part of the same object as p[0]<\/code><\/strong>, so reading memory in between \/ out that far is allowed and won’t segfault.  e.g. allowing auto-vectorization of a search loop.<\/p>\n
But I doubt that gcc even looks for that, let alone takes advantage of it.<\/p>\n
(Note that your question title was specific to gcc targeting x86-64 on Linux, not about whether similar things are safe<\/strong> in gcc, e.g. if done in separate statements.  I mean yes probably safe in practice, but won’t be optimized away almost immediately after parsing.  And definitely not about C++ in general.)<\/p>\n
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I highly recommend not<\/em> doing this.  Use uintptr_t<\/code> to hold pointer-like values that aren’t actual valid pointers.<\/strong>  like you’re doing in the updates to your answer on C++ gcc extension for non-zero-based array pointer allocation?.<\/p>\n<\/p><\/div>\n
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            10<\/span> <\/p><\/div>\n<\/div>\n
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solved Does (p+x)-x always result in p for pointer p and integer x in gcc linux x86-64 C++ <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Yes, for gcc5.x and later specifically, that specific expression is optimized very early to just p, even with optimization disabled, regardless of any possible runtime UB. This happens even with a static array and compile-time constant size. gcc -fsanitize=undefined doesn’t insert any instrumentation to look for it either. Also no warnings at -Wall -Wextra … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[324,608,424,1035,1034],"class_list":["post-4712","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c","tag-gcc","tag-linux","tag-pointer-arithmetic","tag-x86-64"],"yoast_head":"\n[Solved] Does (p+x)-x always result in p for pointer p and integer x in gcc linux x86-64 C++ - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-does-px-x-always-result-in-p-for-pointer-p-and-integer-x-in-gcc-linux-x86-64-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Does (p+x)-x always result in p for pointer p and integer x in gcc linux x86-64 C++ - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Yes, for gcc5.x and later specifically, that specific expression is optimized very early to just p, even with optimization disabled, regardless of any possible runtime UB. 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