{"id":14322,"date":"2022-10-07T08:44:14","date_gmt":"2022-10-07T03:14:14","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-im-just-beginning-to-learn-c-can-someone-explain-what-the-pointers-and-typecasting-are-doing-in-this-code-closed\/"},"modified":"2022-10-07T08:44:14","modified_gmt":"2022-10-07T03:14:14","slug":"solved-im-just-beginning-to-learn-c-can-someone-explain-what-the-pointers-and-typecasting-are-doing-in-this-code-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-im-just-beginning-to-learn-c-can-someone-explain-what-the-pointers-and-typecasting-are-doing-in-this-code-closed\/","title":{"rendered":"[Solved] I’m just beginning to learn C: Can someone explain what the pointers and typecasting are doing in this code? [closed]"},"content":{"rendered":"

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The snippet<\/p>\n

char *encrypt(char *string, size_t length) {\n}\n<\/code><\/pre>\n
defines<\/em> a function named encrypt<\/code>, which takes a char *<\/code> and a size_t<\/code> as arguments, and returns a char *<\/code>.  As written, the function body is empty and will trigger at least one diagnostic because it isn’t returning a value.  <\/p>\n
It looks like it’s meant to take an unencrypted string as the string<\/code> argument, and return the encrypted version of it:<\/p>\n
char unenc[] = \"This is a test\";\nchar *enc = encrypt( unenc, strlen( unenc ) );\n<\/code><\/pre>\n
We’re dealing with pointers because of how C treats array expressions.  A string<\/em> is a sequence of character values including a 0-valued terminator – the string “hello” is represented as the sequence {'h', 'e', 'l', 'l', 'o', 0}<\/code>.  Strings are stored in arrays of character type (char<\/code> for ASCII, EBCDIC, and UTF-8 encodings, wchar_t<\/code> for “wide” encodings like UTF-16).  <\/p>\n
However, unless it is the operand of the sizeof<\/code> or unary &<\/code> operators, or is a string literal<\/em> used to initialize a character array in a declaration, an expression<\/em> of array type is converted (“decays”) to an expression of pointer type, and the value of the expression is the address of the first element of the array.<\/p>\n
We defined unenc<\/code> as an array of char<\/code> large enough to hold the string \"This is a test\"<\/code> (14 characters plus the 0 terminator).  When we pass unenc<\/code> as an argument to encrypt<\/code>, it “decays” from type char [15]<\/code> to char *<\/code>, and what the function actually receives is the address of the first element of the array (&unenc[0]<\/code>).  <\/p>\n
Array expressions cannot be the target of an assignment, and functions cannot return array types.  However encrypt<\/code> is meant to work internally, it cannot return the encrypted string as an array<\/em>; it can only return a pointer to the first element of the array storing the string.  <\/p>\n
Now, without knowing the implementation, it’s possible that encrypt<\/code> is encrypting the string in place<\/em>, overwriting the original contents of the array, and return that same address:<\/p>\n
char *encrypt( char *string, size_t length )\n{\n  for ( size_t i = 0; i < length; i++ )\n    string[i] = replace_char( string[i] );\n  return string;\n}\n<\/code><\/pre>\n
Since string<\/code> is declared char *<\/code> and not const char *<\/code>, that’s a reasonable inference to make.  And it would avoid memory management headaches.  Of course you’re destroying your input, so there are tradeoffs there.  <\/p>\n
If it’s not encrypting in place, then the encrypt<\/code> function needs to either allocate new memory, or use memory that’s available outside the lifetime of the function (either a global or a local that’s been declared static<\/code>, which brings its own headaches):<\/p>\n
char *encrypt( char *string, size_t length )\n{\n  char *e = malloc( length + 1 );\n  if ( e )\n  {\n    for ( size_t i = 0; i < length; i++ )\n      e[i] = replace_char( string[i] );\n  }\n  return e;\n}\n<\/code><\/pre>\n<\/p><\/div>\n
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solved I’m just beginning to learn C: Can someone explain what the pointers and typecasting are doing in this code? [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] The snippet char *encrypt(char *string, size_t length) { } defines a function named encrypt, which takes a char * and a size_t as arguments, and returns a char *. As written, the function body is empty and will trigger at least one diagnostic because it isn’t returning a value. It looks like it’s meant … 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":[324,712],"class_list":["post-14322","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c","tag-pointers"],"yoast_head":"\n[Solved] I'm just beginning to learn C: Can someone explain what the pointers and typecasting are doing in this code? 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