{"id":15003,"date":"2022-10-10T01:37:53","date_gmt":"2022-10-09T20:07:53","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-advanced-pointer-typecasting-in-c-c-closed\/"},"modified":"2022-10-10T01:37:53","modified_gmt":"2022-10-09T20:07:53","slug":"solved-advanced-pointer-typecasting-in-c-c-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-advanced-pointer-typecasting-in-c-c-closed\/","title":{"rendered":"[Solved] Advanced pointer typecasting in C\/C++ [closed]"},"content":{"rendered":"

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The memory simply contains bytes of data. It is up to you how you interpret this data.<\/p>\n

Your buffer may be interpreted as int<\/code>. For example:<\/p>\n

char buffer[] = {1, 2, 3, 4};\nint number = *(int *)buffer;\n<\/code><\/pre>\n
Running the code on an Intel x86 processor would result in number<\/code> having the value of 0x04030201<\/code>, as integers are stored as 32bit (4 bytes) little-endian buffers.<\/p>\n
Your buffer can also be interpreted as code. The C code return 5;<\/code> may be compiled into the following x86 assembly code:<\/p>\n
mov eax, 5\nretn\n<\/code><\/pre>\n
This code looks in the memory as B8 05 00 00 00 C3<\/code><\/p>\n
So let’s look on the following example:<\/p>\n
char buffer[] = {0xb8, 5, 0, 0, 0, 0xc3};\nint number = ((int (*)())(buffer))();\n<\/code><\/pre>\n
Running this code on an Intel x86 processor (assuming you compile this code without security checks that prevent the execution of data buffers) would result in number<\/code> having the value of 5; The code casts the pointer to buffer<\/code> into a pointer to a function with the signature int func_name()<\/code> and then executes it.<\/p>\n<\/p><\/div>\n
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            1<\/span> <\/p><\/div>\n<\/div>\n
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solved Advanced pointer typecasting in C\/C++ [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] The memory simply contains bytes of data. It is up to you how you interpret this data. Your buffer may be interpreted as int. For example: char buffer[] = {1, 2, 3, 4}; int number = *(int *)buffer; Running the code on an Intel x86 processor would result in number having the value of … 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":[465,324,387,712],"class_list":["post-15003","post","type-post","status-publish","format-standard","hentry","category-solved","tag-assembly","tag-c","tag-casting","tag-pointers"],"yoast_head":"\n[Solved] Advanced pointer typecasting in C\/C++ [closed] - 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-advanced-pointer-typecasting-in-c-c-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Advanced pointer typecasting in C\/C++ [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] The memory simply contains bytes of data. It is up to you how you interpret this data. Your buffer may be interpreted as int. For example: char buffer[] = {1, 2, 3, 4}; int number = *(int *)buffer; Running the code on an Intel x86 processor would result in number having the value of ... 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