{"id":19754,"date":"2022-11-07T16:04:49","date_gmt":"2022-11-07T10:34:49","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-what-is-the-output-please-explain-considering-i-am-a-novice-in-c-closed\/"},"modified":"2022-11-07T16:04:49","modified_gmt":"2022-11-07T10:34:49","slug":"solved-what-is-the-output-please-explain-considering-i-am-a-novice-in-c-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-what-is-the-output-please-explain-considering-i-am-a-novice-in-c-closed\/","title":{"rendered":"[Solved] what is the output? Please explain, considering i am a novice in c [closed]"},"content":{"rendered":"

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Time for a crash course on arrays in C.<\/p>\n

First of all, let’s fix the initializer for the array:<\/p>\n

int a[3][4] = {\n                { 1,  2,  3,  4},\n                { 5,  6,  7,  8},\n                { 9, 10, 11, 12}\n              };\n<\/code><\/pre>\n
This defines a 3-element array of 4-element arrays of int<\/code>.  The type of the expression<\/em> a<\/code> is “3-element array of 4-element arrays of int<\/code>“.   <\/p>\n
Now for the headache-inducing part.  Except when it’s the operand of the sizeof<\/code> or unary &<\/code> operators, or if it’s a string literal being used to initialize another array in a declaration, an expression of array type will have its type implicitly converted (“decay”) to a pointer type.  <\/p>\n
If the expression a<\/code> appears by itself in the code (such as in a statement like printf(\"%p\", a);<\/code>, its type is converted from “3-element array of 4-element array of int<\/code>” to “pointer to 4-element array of int<\/code>“, or int (*)[4]<\/code>.  Similarly, if the expression a[i]<\/code> appears in the code, its type is converted from “4-element array of int<\/code>” (int [4]<\/code>) to “pointer to int<\/code>” (int *<\/code>).  If a<\/code> or a[i]<\/code> are operands of either sizeof<\/code> or &<\/code>, however, the conversion doesn’t happen.  <\/p>\n
In a similar vein, array subscripting is done through pointer arithmetic: the expression a[i]<\/code> is interpreted as though it were written *(a+i)<\/code>.  You offset i<\/code> elements from the base of the array and dereference the result.  Thus, a[0]<\/code> is the same as *(a + 0)<\/code>, which is the same as *a<\/code>.  a[i][j]<\/code> is the same as writing *(*(a + i) + j)<\/code>.  <\/p>\n
Here’s a table summarizing all of the above:<\/p>\n
\nExpression   Type          Decays To    Resulting Value\n----------   ----          ---------    -----\n         a   int [3][4]    int (*)[4]   Address of the first element of the array\n        &a   int (*)[3][4] n\/a          Same as above, but type is different\n        *a   int [4]       int *        Same as above, but type is different\n      a[0]   int [4]       int *        Same as above\n    *(a+0)   int [4]       int *        Same as above\n      a[i]   int [4]       int *        Address of the first element of the i'th subarray\n    *(a+i)   int [4]       int *        Same as above\n     &a[i]   int (*)[4]    n\/a          Same as above, but type is different\n     *a[i]   int           n\/a          Value of the 0'th element of the i'th subarray\n   a[i][j]   int                        Value of the j'th element of the i'th subarray\n *(a[i]+j)   int                        Same as above\n*(*(a+i)+j)  int                        Same as above\n<\/pre>\n
Hopefully, that should give you everything you need to figure out what the output should be.  However, the printf<\/code> statement should be written as<\/p>\n
printf(\"%p %d %d\\n\", (void *) a[0]+1, *(a[0]+1), *(*(a+0)+1));\n<\/code><\/pre>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            1<\/span> <\/p><\/div>\n<\/div>\n
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solved what is the output? Please explain, considering i am a novice in c [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Time for a crash course on arrays in C. First of all, let’s fix the initializer for the array: int a[3][4] = { { 1, 2, 3, 4}, { 5, 6, 7, 8}, { 9, 10, 11, 12} }; This defines a 3-element array of 4-element arrays of int. The type of the expression … 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":[361,324,505,712],"class_list":["post-19754","post","type-post","status-publish","format-standard","hentry","category-solved","tag-arrays","tag-c","tag-multidimensional-array","tag-pointers"],"yoast_head":"\n[Solved] what is the output? 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