指向指针指针的指针问题

Masood / ******************************* ******************* **** / #include< stdio.h> #include < stdlib.h> / * #define INLINE_INIT * / #define MAXROWS 3 ＃定义MAXCOLS 5 void buildTbl（int *** tblPtr，size_t numRows，size_t numCols） { * tblPtr =（int **）malloc（numRows * sizeof（int *））; / * C ++：* tblPtr = new（int *）[numRows]; * / for（size_t i = 0; i< numCols; i ++） * tblPtr [i] =（int *）malloc（numCols * sizeof（int））; / * C ++：* tblPtr [i] = new（int）[numCols]; * / } main（） { int startVal = 5; int ** tbl; #ifdef INLINE_INIT tbl =（int **）malloc（MAXROWS * sizeof（int *） ）; / * C ++：tbl = new（int *）[MAXROWS]; * / for（size_t i = 0; i< MAXCOLS; i ++） tbl [i] =（int *）malloc（MAXCOLS * sizeof （int））; / * C ++：tbl [i] = new（int）[MAXCOLS]; * / #else buildTbl（& tbl，MAXROWS，MAXCOLS）; #endif for（size_t row = 0; row< MAXROWS; row ++） for（size_t col = 0; col< MAXCOLS; col ++） tbl [row ] [col] = startVal ++; for（size_t row = 0; row< MAXROWS; row ++） for（size_t col = 0; col< ; MAXCOLS; col ++） printf（" Row：％d，Col：％d =>％d \ n"， row，col，tbl [行] [col]）; 返回0; } 解决方案 ma**********@lycos.com 写道： Masood / ******************************* ******************* **** / #include< stdio.h> #include< stdlib.h> / * #define INLINE_INIT * / #define MAXROWS 3 #define MAXCOLS 5 void buildTbl（int * ** tblPtr，size_t numRows，size_t numCols） { * tblPtr =（int **）malloc（numRows * sizeof（int *））; / * C ++：* tblPtr = new（int *）[numRows]; * / for（size_t i = 0; i< numCols; i ++） * tblPtr [i] =（int *）malloc（numCols * sizeof（int））; （* tblPtr）[i] = ... 注意x = malloc（n * sizeof * x）; 优先于x =（type *）malloc（n * sizeof（type）; 你应该检查malloc是否没有返回NULL。 你可以将buildTbl的返回类型更改为int **和 返回数组而不是通过参数。 我还记得吗东西好吗？很高兴看到你们大部分都是 还在这里！ Tobias / * C ++：* tblPtr [i ] = new（int）[numCols]; * / } main（） {int / startVal = 5; int ** tbl; #ifdef INLINE_INIT tbl =（int **）malloc（MAXROWS * sizeof（int *））; / * C ++：tbl = new（ int *）[MAXROWS]; * / for（size_t i = 0; i< MAXCOLS; i ++） tbl [i] =（int *）malloc（MAXCOLS * sizeof（ int））; / * C ++：tbl [i] = new（int）[MAXCOLS]; * / #else buildTbl（& tbl，MAXROWS，MAXCOLS）; #endif for（size_t row = 0; r ow< MAXROWS;行++） for（size_t col = 0; col< MAXCOLS; col ++） tbl [row] [col] = startVal ++; for（size_t row = 0; row < MAXROWS; row ++） for（size_t col = 0; col< MAXCOLS; col ++） printf（" Row：％d，Col：％d =>％d \ n" ， row，col，tbl [row] [col]）; 返回0; } ma**********@lycos.com 写道：下面的代码示例显示了2D数组的动态分配。我必须承认，我花了很长时间才到达那里（我已经有了另一个帖子），但我很高兴我终于得到了它的工作。现在问题就在于： 我能够正确地动态分配2D阵列，因为我正在正在进行。 （即没有调用任何功能）。当我尝试在另一个函数中执行它时，我得到了一个核心转储。任何帮助将不胜感激。 由于此函数需要更新指针指针类型，我实际上是将指针传递给指针指针类型为函数。我在这里错过了什么？ 你可以迷失在C的错综复杂中。 一个诡计绕过它：使用本地指向指针类型> 变量来执行常规操作并将其值分配给对象 your your指针到指针指针类型论点指向。 - >基本相同的代码，一个额外的行。 （你在* t​​blPtr上执行所有操作，这实际上意味着 你也可以使用int ** tbl并放* tblPtr = tbl在最后 如果一切顺利的话） 如果你遵循Tobias的建议，那就更容易了。 你我可以看到源代码在我执行二维数组初始化在线时正常工作。 （即不通过调用函数调用）只需取消注释该行 / * #define INLINE_INIT * / Masood [snip：代码] Tobias Oed已经提到了我能看到的一切。 关注他关于malloc的建议的每一点（ ）。 BTW：谢谢你提供一个很好的例子！ 干杯 Michael - 电子邮件：我的是/ at / gmx / dot / de地址。 ma ********** @ lycos.com 写道： void buildTbl（int *** tblPtr，size_t numRows，size_t numCols） { * tblPtr =（int **）malloc（numRows * sizeof（int *））; / * C ++：* tblPtr = new（int *）[numRows]; * / for（size_t i = 0; i< numCols; i ++） 条件应为i< numRows。 * tblPtr [i] =（int *）malloc（numCols * sizeof（int））; 这将访问内存如果i大于0，则不属于tblPtr。 将其更改为以下内容： （* tblPtr）[i] = malloc（numCols * sizeof（int）） ; 检查malloc（）的返回值也是个好主意。 Christian The code example below shows the dynamic allocation of a 2D array. Imust admit that it took quite a while for me to get there (I alreadyhave another posting to that effect), but I am glad that I finally gotit working. Now here''s the problem:I am able to get the 2D array dynamically allocated correctly as longas I am doing it "in-line" (i.e. without invoking any function). Themoment I try to do it in another function, I get a core dump. Any helpwill be appreciated.Since this function is expected to update a pointer-to-pointer type, Iam actually passing the pointer-to-pointer-to-pointer type to thefunction. What am I missing here?You can see that the source code works correctly when I am perform 2Darray initialization "in-line" (i.e. by not invoking a functioncall) simply by un-commenting the line/* #define INLINE_INIT */Masood/************************************************** ****/#include <stdio.h>#include <stdlib.h>/*#define INLINE_INIT*/#define MAXROWS 3#define MAXCOLS 5voidbuildTbl(int ***tblPtr, size_t numRows, size_t numCols){*tblPtr = (int **)malloc(numRows*sizeof(int*));/* C++ : *tblPtr = new (int*)[numRows]; */for(size_t i = 0; i < numCols; i++)*tblPtr[i] = (int *)malloc(numCols*sizeof(int));/* C++: *tblPtr[i] = new (int)[numCols]; */}main(){int startVal = 5;int **tbl;#ifdef INLINE_INITtbl = (int **)malloc(MAXROWS*sizeof(int*));/* C++ : tbl = new (int*)[MAXROWS]; */for(size_t i = 0; i < MAXCOLS; i++)tbl[i] = (int *)malloc(MAXCOLS*sizeof(int));/* C++: tbl[i] = new (int)[MAXCOLS]; */#elsebuildTbl(&tbl, MAXROWS, MAXCOLS);#endiffor(size_t row = 0; row < MAXROWS; row++)for(size_t col = 0; col < MAXCOLS; col++)tbl[row][col] = startVal++;for(size_t row = 0; row < MAXROWS; row++)for(size_t col = 0; col < MAXCOLS; col++)printf("Row: %d, Col: %d => %d\n",row, col, tbl[row][col]);return 0;} 解决方案 ma**********@lycos.com wrote: Masood /************************************************** ****/ #include <stdio.h> #include <stdlib.h> /*#define INLINE_INIT*/ #define MAXROWS 3 #define MAXCOLS 5 void buildTbl(int ***tblPtr, size_t numRows, size_t numCols) { *tblPtr = (int **)malloc(numRows*sizeof(int*)); /* C++ : *tblPtr = new (int*)[numRows]; */ for(size_t i = 0; i < numCols; i++) *tblPtr[i] = (int *)malloc(numCols*sizeof(int));(*tblPtr)[i] = ...Note that x = malloc(n * sizeof *x);is preferred to x = (type *) malloc(n * sizeof(type);You should check that malloc doesn''t return NULL.You could change the return type of buildTbl to int ** andreturn the array that way instead of through the parameter.Do I remember that stuff Ok? Nice to see most of you arestill here!Tobias /* C++: *tblPtr[i] = new (int)[numCols]; */ } main() { int startVal = 5; int **tbl; #ifdef INLINE_INIT tbl = (int **)malloc(MAXROWS*sizeof(int*)); /* C++ : tbl = new (int*)[MAXROWS]; */ for(size_t i = 0; i < MAXCOLS; i++) tbl[i] = (int *)malloc(MAXCOLS*sizeof(int)); /* C++: tbl[i] = new (int)[MAXCOLS]; */ #else buildTbl(&tbl, MAXROWS, MAXCOLS); #endif for(size_t row = 0; row < MAXROWS; row++) for(size_t col = 0; col < MAXCOLS; col++) tbl[row][col] = startVal++; for(size_t row = 0; row < MAXROWS; row++) for(size_t col = 0; col < MAXCOLS; col++) printf("Row: %d, Col: %d => %d\n", row, col, tbl[row][col]); return 0; } ma**********@lycos.com wrote: The code example below shows the dynamic allocation of a 2D array. I must admit that it took quite a while for me to get there (I already have another posting to that effect), but I am glad that I finally got it working. Now here''s the problem: I am able to get the 2D array dynamically allocated correctly as long as I am doing it "in-line" (i.e. without invoking any function). The moment I try to do it in another function, I get a core dump. Any help will be appreciated. Since this function is expected to update a pointer-to-pointer type, I am actually passing the pointer-to-pointer-to-pointer type to the function. What am I missing here?You can get lost with the intricacies of C.One "trick" to get around it: Use a local "pointer-to-pointer type"variable to do as you always did and assign its value to the objectyour "pointer-to-pointer-to-pointer type" argument points to.-> Essentially the same code, one additional line.(You perform all operations on *tblPtr which essentially means thatyou could also work with a int **tbl and put *tblPtr=tbl at the endif everything worked out)If you follow Tobias''s advice, it becomes even easier. You can see that the source code works correctly when I am perform 2D array initialization "in-line" (i.e. by not invoking a function call) simply by un-commenting the line /* #define INLINE_INIT */ Masood[snip: Code]Tobias Oed has already mentioned everything I can see.Follow _every_ point of his advice regarding malloc().BTW: Thank you for providing a good minimal example!CheersMichael--E-Mail: Mine is an /at/ gmx /dot/ de address.ma**********@lycos.com wrote: void buildTbl(int ***tblPtr, size_t numRows, size_t numCols) { *tblPtr = (int **)malloc(numRows*sizeof(int*)); /* C++ : *tblPtr = new (int*)[numRows]; */ for(size_t i = 0; i < numCols; i++)The condition should be i < numRows. *tblPtr[i] = (int *)malloc(numCols*sizeof(int));This will access memory not belonging to tblPtr if i is greater than 0.Change it to the following:(*tblPtr)[i] = malloc(numCols*sizeof(int));It''s also a good idea to check the return value of malloc().Christian 这篇关于指向指针指针的指针问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！