问题描述
我想通过引用将我的二维字符数组传递给另一个函数。我用我的 int ** 做了同样的事情，它工作得很好。但是当谈到 char ** 时，同样的方法被打破了。我尝试了太多东西。我几乎阅读了所有问题，但我看不到我的案例。
首先，我有一个设置为 NULL 的单字符 **，我通过引用将它传递给另一个函数。
这是我的主要功能
int main(){
int **recommendU2 = NULL;
char **bookNames = NULL;
readFromCsv(&recommendU2,&bookNames,&bookCount,&userCount);
printf("\n user[2][3] = %d \n",recommendU2[2][3]); //this works
printf("\n bookNames[1] = %s \n,bookNames[1]); //this doesn't works
return 0;
}
这是我的另一个函数，它从 csv 读取并处理这些数据。它是做它的目的没有和问题。它从 csv 读取并逐行读取数据并毫无问题地进行处理。但我无法访问我的 char ** 数据。但是我可以访问我的 int ** 数据。我也没发现有什么区别。
void readFromCsv(int ***user,char ***bookNames,int *bookCount,int *userCount){
FILE * fpointer;
int lineCount = 0;
int count;
int sizes[3];
fpointer = fopen("recdataset.csv","r");
char recommendationSheet[RECOMMENDATIONSIZE];
while(fgets(recommendationSheet,RECOMMENDATIONSIZE,fpointer)){
if(lineCount>1){
//this part is not important for this question
}
else if(lineCount == 0){
userArraysMemoryAllocate(recommendationSheet,user,nuser,bookNames,&sizes[0]);
(*user)[2][3] = 4;
//I can access this user data both within this function and main function
}
else if(lineCount == 1){
//actually I send variables to another function which named getBookNames() but when I can't reach data I SUSPECT it and move it's code to here. But this doesn't make any different too.
char *p = recommendationSheet;
int j = 0;
int i = 0;
while(1) {
char *p2 = strchr(p,';');
if(p2 != NULL)
*p2 = '\0';
if(i>0)
strcpy((*bookNames)[i],p);
printf("\n Book Name is = %s \n",(*bookNames)[i]); //This line works read bookNames correctly
i++;
if(p2 == NULL)
break;
p = p2 + 1;
}
printf("\nbookName = %s \n",(*bookNames)[2]); //this line doesn't work
}
lineCount++;
}
fclose(fpointer);
}
我无法让它工作。我尝试了很多变体，但没有一个有效。如果您有任何解决方案或任何其他想法来完成这项工作，我将不胜感激。提前致谢。
注意：这个程序太长，放在这里。所以我只在这里复制了有问题的部分。这包括如此多的函数调用。所以如果你有不明白的地方请说出来，我也可以添加这部分代码。
更新
我也向 readfromCsv 函数添加了内存分配函数。这是代码。我的问题仍然存在。我以为我修复了它，但它导致了异常行为。
void readFromCsv(int ***user,int *userCount,){
FILE * fpointer;
int lineCount = 0;
int i = 0;
int j = 0;
int len;
int count;
int sizes[3];
fpointer = fopen("recdataset.csv","r");
char recommendationSheet[RECOMMENDATIONSIZE];
while(fgets(recommendationSheet,fpointer)){
len = strlen(recommendationSheet);
if( recommendationSheet[len-1] == '\n' )
recommendationSheet[len-1] = 0;
if(lineCount>1){
//not important
}
else if(lineCount == 0){
char *p = recommendationSheet;
i = 0;
j = 0;
int row = 0;
int col = 0;
while(i<3) {
char *p2 = strchr(p,';');
if(p2 != NULL)
*p2 = '\0';
if(i==0){
row = atoi(p);
col = row;
if((*bookNames = (char**)malloc(row * sizeof(char*))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
for(j = 0; j < row; ++j){
if(((*bookNames)[j] = (char*)malloc(50 * sizeof(char))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
}
}
else if(i==1){
row = atoi(p);
if((*user = (int**)malloc(row * sizeof(int*))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
for(j = 0; j < row; ++j){
if(((*user)[j] = (int*)malloc(col * sizeof(int))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
}
}
else if(i==2){
row = atoi(p);
if((*nuser = (int**)malloc(row * sizeof(int*))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
for(j = 0; j < row; ++j){
if(((*nuser)[j] = (int*)malloc(col * sizeof(int))) == NULL){
printf("\nMemory allocation problem \n");
exit(0);
}
}
}
sizes[i] = row;
i++;
if(p2 == NULL)
break;
p = p2 + 1;
}
(*user)[2][3] = 4; // this works still here and main function
}
else if(lineCount == 1){
char *pMemory = recommendationSheet;
j = 0;
i = 0;
while(1) {
char *p2 = strchr(pMemory,';');
if(p2 != NULL)
*p2 = '\0';
if(i>0)
(*bookNames)[i] = pMemory;
printf("\n %d. Book name is = %s \n",i,(*bookNames)[i]); // works perfectly
i++;
if(p2 == NULL)
break;
pMemory = p2 + 1;
}
}
lineCount++;
}
fclose(fpointer);
}
我仍然无法从主函数访问我的 char **。我添加了内存分配过程。我虽然修复了它，但问题再次发生而没有任何变化。我根本不明白这个问题。
解决方法
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