如何解决如何在C中返回具有动态多维数组的结构？
我遇到以下问题：“如果尝试返回具有动态多维数组的结构，则会丢失该数组内部的数据。如果使用指针返回该结构，则会遇到相同的问题。在某些情况下，我会遇到同样的问题。在数组中具有预期数据的位置” ->如何返回包含动态多维数组的结构？ ->我的错误在哪里？
以下是显示问题的示例程序：
#include <stdio.h>
#include <stdlib.h>
struct just_a_structure
{
double **lf_multi_array;
float *f_array;
};
struct just_a_structure get_struct(int i_length)
{
struct just_a_structure return_this_struct;
return_this_struct.lf_multi_array = (double**)malloc(i_length*sizeof(double*));
return_this_struct.f_array = (float*)malloc(i_length*sizeof(float*));
double lf_dummy_0[10] = {0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9};
double lf_dummy_1[10] = {1.0,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9};
return_this_struct.lf_multi_array[0] = lf_dummy_0;
return_this_struct.f_array[0] = 0.00;
return_this_struct.lf_multi_array[1] = lf_dummy_1;
return_this_struct.f_array[1] = 0.01;
printf("\n output in function \n");
printf("ma[0][0] %f \n",return_this_struct.lf_multi_array[0][0]);
printf("ma[0][1] %f \n",return_this_struct.lf_multi_array[0][1]);
printf("ma[0][2] %f \n",return_this_struct.lf_multi_array[0][2]);
printf("ma[0][3] %f \n",return_this_struct.lf_multi_array[0][3]);
printf("ma[0][4] %f \n",return_this_struct.lf_multi_array[0][4]);
printf("ma[0][5] %f \n",return_this_struct.lf_multi_array[0][5]);
printf("ma[0][6] %f \n",return_this_struct.lf_multi_array[0][6]);
printf("ma[0][7] %f \n",return_this_struct.lf_multi_array[0][7]);
printf("ma[0][8] %f \n",return_this_struct.lf_multi_array[0][8]);
printf("ma[0][9] %f \n",return_this_struct.lf_multi_array[0][9]);
printf("a[0] %f \n",return_this_struct.f_array[0]);
printf("\n \n");
printf("ma[1][0] %f \n",return_this_struct.lf_multi_array[1][0]);
printf("ma[1][1] %f \n",return_this_struct.lf_multi_array[1][1]);
printf("ma[1][2] %f \n",return_this_struct.lf_multi_array[1][2]);
printf("ma[1][3] %f \n",return_this_struct.lf_multi_array[1][3]);
printf("ma[1][4] %f \n",return_this_struct.lf_multi_array[1][4]);
printf("ma[1][5] %f \n",return_this_struct.lf_multi_array[1][5]);
printf("ma[1][6] %f \n",return_this_struct.lf_multi_array[1][6]);
printf("ma[1][7] %f \n",return_this_struct.lf_multi_array[1][7]);
printf("ma[1][8] %f \n",return_this_struct.lf_multi_array[1][8]);
printf("ma[1][9] %f \n",return_this_struct.lf_multi_array[1][9]);
printf("a[1] %f \n",return_this_struct.f_array[1]);
return return_this_struct;
}
int main()
{
int i_length = 500;
struct just_a_structure returned_struct;
returned_struct = get_struct(i_length);
printf("\n output in main \n");
printf("ma[0][0] %f \n",returned_struct.lf_multi_array[0][0]);
printf("ma[0][1] %f \n",returned_struct.lf_multi_array[0][1]);
printf("ma[0][2] %f \n",returned_struct.lf_multi_array[0][2]);
printf("ma[0][3] %f \n",returned_struct.lf_multi_array[0][3]);
printf("ma[0][4] %f \n",returned_struct.lf_multi_array[0][4]);
printf("ma[0][5] %f \n",returned_struct.lf_multi_array[0][5]);
printf("ma[0][6] %f \n",returned_struct.lf_multi_array[0][6]);
printf("ma[0][7] %f \n",returned_struct.lf_multi_array[0][7]);
printf("ma[0][8] %f \n",returned_struct.lf_multi_array[0][8]);
printf("ma[0][9] %f \n",returned_struct.lf_multi_array[0][9]);
printf("a[0] %f \n",returned_struct.f_array[0]);
printf("\n \n");
printf("ma[1][0] %f \n",returned_struct.lf_multi_array[1][0]);
printf("ma[1][1] %f \n",returned_struct.lf_multi_array[1][1]);
printf("ma[1][2] %f \n",returned_struct.lf_multi_array[1][2]);
printf("ma[1][3] %f \n",returned_struct.lf_multi_array[1][3]);
printf("ma[1][4] %f \n",returned_struct.lf_multi_array[1][4]);
printf("ma[1][5] %f \n",returned_struct.lf_multi_array[1][5]);
printf("ma[1][6] %f \n",returned_struct.lf_multi_array[1][6]);
printf("ma[1][7] %f \n",returned_struct.lf_multi_array[1][7]);
printf("ma[1][8] %f \n",returned_struct.lf_multi_array[1][8]);
printf("ma[1][9] %f \n",returned_struct.lf_multi_array[1][9]);
printf("a[1] %f \n",returned_struct.f_array[1]);
return 0;
}
这是程序的输出：
output in function
ma[0][0] 0.000000
ma[0][1] 0.100000
ma[0][2] 0.200000
ma[0][3] 0.300000
ma[0][4] 0.400000
ma[0][5] 0.500000
ma[0][6] 0.600000
ma[0][7] 0.700000
ma[0][8] 0.800000
ma[0][9] 0.900000
a[0] 0.000000
ma[1][0] 1.000000
ma[1][1] 1.100000
ma[1][2] 1.200000
ma[1][3] 1.300000
ma[1][4] 1.400000
ma[1][5] 1.500000
ma[1][6] 1.600000
ma[1][7] 1.700000
ma[1][8] 1.800000
ma[1][9] 1.900000
a[1] 0.010000
output in main
ma[0][0] 27471811305989270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][1] 27471811305989273000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][2] 24361411570075108000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][3] 22932960320429427000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][4] 0.000000
ma[0][5] 0.000000
ma[0][6] 23754367043167420000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][7] 27471811305989273000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][8] -519591837366427200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[0][9] 0.000000
a[0] 0.000000
ma[1][0] 21266032917854245000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[1][1] 21427427439475540000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[1][2] 0.000000
ma[1][3] 0.000000
ma[1][4] 0.000000
ma[1][5] 0.000000
ma[1][6] 22334679298120726000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[1][7] -1.#QNAN0
ma[1][8] 23669996832223273000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
ma[1][9] 0.000000
a[1] 0.010000
我试图实现@Etini和@Sourav Ghost提供的解决方案。但是知道程序启动后立即崩溃。
return_this_struct.lf_multi_array[0] = (double*)malloc(10*sizeof(double));
return_this_struct.lf_multi_array[1] = (double*)malloc(10*sizeof(double));
return_this_struct.f_array = (float*)malloc(i_length*sizeof(float*));
double lf_dummy_0[10] = {0.0,1.9};
for(int i = 0; i < 10; i++)
{
return_this_struct.lf_multi_array[0][i] = lf_dummy_0[i];
}
for(int i = 0; i < 10; i++)
{
return_this_struct.lf_multi_array[1][i] = lf_dummy_1[i];
}
return_this_struct.f_array[0] = 0.00;
return_this_struct.f_array[1] = 0.01;
```
解决方法
在您的代码中
return_this_struct.lf_multi_array[0] = lf_dummy_0;
和
return_this_struct.lf_multi_array[1] = lf_dummy_1;
两者都在指针中存储了局部变量的地址，因此，一旦从被调用函数返回，这些局部变量就会超出范围，并且您正在访问无效的内存，从而产生不确定的行为。
如果要返回值，请为数组分配内存，然后复制单个元素值。
,正如Sourav Ghost所说，您需要分配和复制单个元素，这就是我使它适用于第一个数组的方式
return_this_struct.lf_multi_array[0] = (double*) malloc(10*sizeof(double));
for(int i = 0; i < 10; i++){
return_this_struct.lf_multi_array[0][i] = lf_dummy_0[i];
}
我的C生锈了，请耐心等待。
这两行正在创建一个堆栈存储器，退出该函数时将删除该堆栈存储器：
double lf_dummy_0[10] = {0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9};
double lf_dummy_1[10] = {1.0,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9};
这两行并未按照您的想法将存储在lf_dummy_0和lf_dummy_0中的值复制到您的multi_array中。
return_this_struct.lf_multi_array[0] = lf_dummy_0;
...
return_this_struct.lf_multi_array[1] = lf_dummy_1;
现在，我们如何将您的问题分解成碎片？
首先，让我们编写一个方法，该方法将返回单个值数组，并确保在函数结束后可以访问这些值。
#include <stdio.h>
#include <stdlib.h>
double *getPointerToArrayWithValues(int capacity){
double * array = (double *) malloc(sizeof(double) * capacity);
double value = 0.0;
for (int i = 0; i < capacity; ++i){
array[i] = value;
value += 0.1;
}
return array;
}
int main()
{
int size = 10;
double * array = getPointerToArrayWithValues(size);
for (int i = 0; i < size; ++i){
printf("arra[%d]: %f\n",i,array[i]);
}
free(array);
return 0;
}
您可能会注意到，在获得值并显示它们之后，我正在释放我们使用的内存。我们必须自己清理。
还请注意，我使用的是double而不是float。原因是，晚于混合使用float和double会更容易使用它。
第二，让我们看看是否可以创建一个或多个数组。多阵列。
#include <stdio.h>
#include <stdlib.h>
double **getPointerToMultiArrayWithValues(int rowCapacity,int columnCapacity){
// Add space for rows
double ** multiArray = (double **) malloc(sizeof(double * ) * rowCapacity);
// Add space for columns
for(int i = 0; i < rowCapacity; ++i){
multiArray[i] = (double *) malloc(sizeof(double *) * columnCapacity);
}
double value = 0.0;
for (int row = 0; row < rowCapacity; ++row){
for (int column = 0; column < columnCapacity; ++column){
multiArray[row][column] = value;
value += 0.1;
}
}
return multiArray;
}
int main()
{
int numberOfArrayToHold = 10;
int sizeOfEachArray = 10;
double ** multiArray = getPointerToMultiArrayWithValues(numberOfArrayToHold,sizeOfEachArray);
for(int i = 0; i < numberOfArrayToHold; ++i){
for (int j = 0; j < sizeOfEachArray; ++j){
printf("multiArray[%d][%d]: %f\n",j,multiArray[i][j]);
}
}
// Lets be grown up and clean after ourselves
for (int i = 0; i < numberOfArrayToHold; ++i){
free(multiArray[i]);
}
free(multiArray);
return 0;
}
现在，您已经解决了如何创建一个数组和一个多数组（在本例中为双数组）的问题，我们可以集中精力创建一个可以容纳该数组的结构：
#include <stdio.h>
#include <stdlib.h>
struct ArrayHolder{
int rows;
int columns;
double ** multiArray;
};
double **getPointerToMultiArrayWithValues(int rowCapacity,int columnCapacity){
// Add space for rows
double ** multiArray = (double **) malloc(sizeof(double * ) * rowCapacity);
// Add space for columns
for(int i = 0; i < rowCapacity; ++i){
multiArray[i] = (double *) malloc(sizeof(double *) * columnCapacity);
}
double value = 0.0;
for (int row = 0; row < rowCapacity; ++row){
for (int column = 0; column < columnCapacity; ++column){
multiArray[row][column] = value;
value += 0.1;
}
}
return multiArray;
}
int main()
{
struct ArrayHolder arrayHolder;
arrayHolder.rows = 10;
arrayHolder.columns = 10;
arrayHolder.multiArray = getPointerToMultiArrayWithValues(arrayHolder.rows,arrayHolder.columns);
for(int i = 0; i < arrayHolder.rows; ++i){
for (int j = 0; j < arrayHolder.columns; ++j){
printf("multiArray[%d][%d]: %f\n",arrayHolder.multiArray[i][j]);
}
}
for (int i = 0; i < arrayHolder.rows; ++i){
free(arrayHolder.multiArray[i]);
}
free( arrayHolder.multiArray);
return 0;
}
但是，我们希望返回一个结构，不是吗？然后，
#include <stdio.h>
#include <stdlib.h>
struct ArrayHolder{
int rows;
int columns;
double ** multiArray;
};
double **getPointerToMultiArrayWithValues(int rowCapacity,int columnCapacity){
// Add space for rows
double ** multiArray = (double **) malloc(sizeof(double * ) * rowCapacity);
// Add space for columns
for(int i = 0; i < rowCapacity; ++i){
multiArray[i] = (double *) malloc(sizeof(double *) * columnCapacity);
}
double value = 0.0;
for (int row = 0; row < rowCapacity; ++row){
for (int column = 0; column < columnCapacity; ++column){
multiArray[row][column] = value;
value += 0.1;
}
}
return multiArray;
}
struct ArrayHolder getStructWithArray(){
struct ArrayHolder arrayHolder;
arrayHolder.rows = 10;
arrayHolder.columns = 10;
arrayHolder.multiArray = getPointerToMultiArrayWithValues(arrayHolder.rows,arrayHolder.columns);
return arrayHolder;
}
int main()
{
struct ArrayHolder arrayHolder = getStructWithArray();
for(int i = 0; i < arrayHolder.rows; ++i){
for (int j = 0; j < arrayHolder.columns; ++j){
printf("multiArray[%d][%d]: %f\n",arrayHolder.multiArray[i][j]);
}
}
for (int i = 0; i < arrayHolder.rows; ++i){
free(arrayHolder.multiArray[i]);
}
free( arrayHolder.multiArray);
return 0;
}
在这种情况下，我们将按值返回结构（复制）
但是，我们也可以通过指针来做到这一点！
#include <stdio.h>
#include <stdlib.h>
struct ArrayHolder{
int rows;
int columns;
double ** multiArray;
};
double **getPointerToMultiArrayWithValues(int rowCapacity,int columnCapacity){
// Add space for rows
double ** multiArray = (double **) malloc(sizeof(double * ) * rowCapacity);
// Add space for columns
for(int i = 0; i < rowCapacity; ++i){
multiArray[i] = (double *) malloc(sizeof(double *) * columnCapacity);
}
double value = 0.0;
for (int row = 0; row < rowCapacity; ++row){
for (int column = 0; column < columnCapacity; ++column){
multiArray[row][column] = value;
value += 0.1;
}
}
return multiArray;
}
struct ArrayHolder * getPointerToStructWithArray(){
struct ArrayHolder* arrayHolder = (struct ArrayHolder *) malloc(sizeof(struct ArrayHolder));
arrayHolder -> rows = 10;
arrayHolder -> columns = 10;
arrayHolder -> multiArray = getPointerToMultiArrayWithValues(arrayHolder -> rows,arrayHolder -> columns);
return arrayHolder;
}
int main()
{
struct ArrayHolder * arrayHolder = getPointerToStructWithArray();
for(int i = 0; i < arrayHolder -> rows; ++i){
for (int j = 0; j < arrayHolder -> columns; ++j){
printf("multiArray[%d][%d]: %f\n",arrayHolder -> multiArray[i][j]);
}
}
for (int i = 0; i < arrayHolder -> rows; ++i){
free(arrayHolder -> multiArray[i]);
}
free( arrayHolder -> multiArray);
free(arrayHolder);
return 0;
}
让我知道这是否对您有所帮助。
,在使用循环分配值之前，您必须像这样分配数组
return_this_struct.lf_multi_array[0] = (double*) malloc(10*sizeof(double));
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