{"id":11401,"date":"2022-09-27T07:13:55","date_gmt":"2022-09-27T01:43:55","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-why-can-i-convert-from-int-to-int-in-the-case-of-the-two-dimensional-array-but-not-in-the-case-of-the-array-of-pointers\/"},"modified":"2022-09-27T07:13:55","modified_gmt":"2022-09-27T01:43:55","slug":"solved-why-can-i-convert-from-int-to-int-in-the-case-of-the-two-dimensional-array-but-not-in-the-case-of-the-array-of-pointers","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-why-can-i-convert-from-int-to-int-in-the-case-of-the-two-dimensional-array-but-not-in-the-case-of-the-array-of-pointers\/","title":{"rendered":"[Solved] Why can I convert from int ** to int * in the case of the two-dimensional array, but not in the case of the array of pointers"},"content":{"rendered":"

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\n<\/p>\n

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You might want to learn how these objects are laid out in memory:<\/p>\n

int array[][3] = { {1,2,3}, {4,5,6}, {7,8,9} };\nint *p_array = (int *) array;\n<\/code><\/pre>\n
This creates an object array<\/code> of type int[][3]<\/code> in memory (in this case stack) physically laid out in a hypothetical (16-bit word size) architecture as:<\/p>\n
  array 0x8000: 0x0001\n        0x8002: 0x0002\n        0x8004: 0x0003\n        0x8006: 0x0004\n        0x8008: 0x0005\n        0x800A: 0x0006\n        0x800C: 0x0007\n        0x800E: 0x0008\n        0x8010: 0x0009\np_array 0x8012: 0x8000\n<\/code><\/pre>\n
Where 0x8000<\/code> denotes the imaginary stack address of array<\/code>.<\/p>\n
However, in your second example:<\/p>\n
int array1[] = { 11, 12 };\nint array2[] = { 13, 14, 15, 16 };\nint array3[] = { 17, 18, 19 };\nint *array_pointers[] = { array1, array2, array3 };\n\nint *array_pointers2 = (int*) array_pointers;\n<\/code><\/pre>\n
Here array_pointers<\/code> is of type int *[]<\/code> which means it is a pointer to array of int<\/code>s. And the memory layout would be like the following:<\/p>\n
array1 0x8000: 0x000B\n       0x8002: 0x000C\narray2 0x8004: 0x000D\n       0x8006: 0x000E\n       0x8008: 0x000F\n       0x800A: 0x0010\narray3 0x800C: 0x0011\n       0x800E: 0x0012\n       0x8010: 0x0013\n<\/code><\/pre>\n
and<\/p>\n
array_pointers:  0x8012: 0x8000 address of array1\n                 0x8014: 0x8004 address of array2\n                 0x8016: 0x800C address of array3\narray_pointers2: 0x8018: 0x8012 address of array_pointers\n<\/code><\/pre>\n
So when you try to print the contents of array_pointers2<\/code>, which now contain pointers to array of int<\/code>s, you will print the address values rather than the numbers they point to.<\/p>\n
Here is an example that outputs the memory address of those objects:<\/p>\n
https:\/\/ideone.com\/T52Ow0<\/a><\/p>\n
Which prints:<\/p>\n
array      0x7ffeda682090: 0x00000000000001\n           0x7ffeda682094: 0x00000000000002\n           0x7ffeda682098: 0x00000000000003\n           0x7ffeda68209c: 0x00000000000004\n           0x7ffeda6820a0: 0x00000000000005\n           0x7ffeda6820a4: 0x00000000000006\n           0x7ffeda6820a8: 0x00000000000007\n           0x7ffeda6820ac: 0x00000000000008\n           0x7ffeda6820b0: 0x00000000000009\np_array    0x7ffeda682038: 0x007ffeda682090\n\narray1     0x7ffeda68204c: 0x0000000000000b\n           0x7ffeda682050: 0x0000000000000c\narray2     0x7ffeda682060: 0x0000000000000d\n           0x7ffeda682064: 0x0000000000000e\n           0x7ffeda682068: 0x0000000000000f\n           0x7ffeda68206c: 0x00000000000010\narray3     0x7ffeda682054: 0x00000000000011\n           0x7ffeda682058: 0x00000000000012\n           0x7ffeda68205c: 0x00000000000013\narray_poin 0x7ffeda682070: 0x007ffeda68204c\n           0x7ffeda682078: 0x007ffeda682060\n           0x7ffeda682080: 0x007ffeda682054\narray_poi2 0x7ffeda682040: 0x007ffeda682070\n<\/code><\/pre>\n
on my computer.<\/p>\n
Source code for that in case external link expires:<\/p>\n
#include <stdio.h>\n#include <string.h>\n\nint main(void)\n{\n    int array[][3] = { {1,2,3}, {4,5,6}, {7,8,9} };\n    int *p_array = (int*) array;\n\n    for (int i = 0; i < 9; i++) {\n        printf(\"%-10s %08p: %016p\\n\", i == 0 ? \"array\" : \"\", &p_array[i], p_array[i]);\n    }\n\n    printf(\"%-10s %08p: %016p\\n\", \"p_array\", &p_array, p_array);\n\n    int array1[] = { 11, 12 };\n    int array2[] = { 13, 14, 15, 16 };\n    int array3[] = { 17, 18, 19 };\n\n    int *array_pointers[] = { array1, array2, array3 };\n    int *array_pointers2 = (int*) array_pointers;\n\n    putchar('\\n');\n\n    for (int i = 0; i < sizeof(array1) \/ sizeof(array1[0]); i++) {\n        printf(\"%-10s %08p: %016p\\n\", i == 0 ? \"array1\" : \"\", &array1[i], array1[i]);\n    }\n\n    for (int i = 0; i < sizeof(array2) \/ sizeof(array2[0]); i++) {\n        printf(\"%-10s %08p: %016p\\n\", i == 0 ? \"array2\" : \"\", &array2[i], array2[i]);\n    }\n\n    for (int i = 0; i < sizeof(array3) \/ sizeof(array3[0]); i++) {\n        printf(\"%-10s %08p: %016p\\n\", i == 0 ? \"array3\" : \"\", &array3[i], array3[i]);\n    }\n\n    for (int i = 0; i < sizeof(array_pointers) \/ sizeof(array_pointers[0]); i++) {\n        printf(\"%-10s %08p: %016p\\n\", i == 0 ? \"array_poin\" : \"\", &array_pointers[i], array_pointers[i]);\n    }\n\n    printf(\"%-10s %08p: %016p\\n\", \"array_poi2\", &array_pointers2, array_pointers2);\n\n    return 0;\n}\n<\/code><\/pre>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            <\/span> <\/p><\/div>\n<\/div>\n
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solved Why can I convert from int ** to int * in the case of the two-dimensional array, but not in the case of the array of pointers <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] You might want to learn how these objects are laid out in memory: int array[][3] = { {1,2,3}, {4,5,6}, {7,8,9} }; int *p_array = (int *) array; This creates an object array of type int[][3] in memory (in this case stack) physically laid out in a hypothetical (16-bit word size) architecture as: array 0x8000: … 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":[361,324,387,652,712],"class_list":["post-11401","post","type-post","status-publish","format-standard","hentry","category-solved","tag-arrays","tag-c","tag-casting","tag-int","tag-pointers"],"yoast_head":"\n[Solved] Why can I convert from int ** to int * in the case of the two-dimensional array, but not in the case of the array of pointers - 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-why-can-i-convert-from-int-to-int-in-the-case-of-the-two-dimensional-array-but-not-in-the-case-of-the-array-of-pointers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Why can I convert from int ** to int * in the case of the two-dimensional array, but not in the case of the array of pointers - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] You might want to learn how these objects are laid out in memory: int array[][3] = { {1,2,3}, {4,5,6}, {7,8,9} }; int *p_array = (int *) array; This creates an object array of type int[][3] in memory (in this case stack) physically laid out in a hypothetical (16-bit word size) architecture as: array 0x8000: ... 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