[英]How to return pointer to array of struct in C
我使用malloc
在 function 中創建了一個student
結構,並用數據填充了它。 我想將結構數組的地址返回給main
並逐行打印,但是在我的實現中,它沒有打印。 我調試了我的代碼,確實,它能夠將我的數組的地址返回給main
。 我不知道為什么它不打印。 有什么想法嗎?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct student Student;
struct student
{
char name[100];
int age;
char sex;
};
Student **getstudents(int n)
{
Student **t = malloc(sizeof *t * n); // memory for the array of pointers
for (int i = 0; i < n; i++) // memory for each individual pointer
{
t[i] = malloc(sizeof **t);
}
/* Data is inputted by user with form <student> <sex> <age>, then get mapped to the struct *t */
return t; /* Pointer to an array of pointers */
}
int main()
{
Student **students;
int n;
scanf("%d\n", &n);
students = getstudents(n);
for (int i = 0; i < n; i++)
{
printf("Name: %s, Sex: %s, Age: %d\n", students[i]->name, students[i]->sex, students[i]->age);
}
for (int i = 0; i < n; i++)
{
free(students[i]);
}
free(students);
return 0;
}
I am only allowed to modify the code in `Student **getstudents(int n)`.
在行中:
Student *t = (char*)malloc(sizeof(Student)*n);
您正在為一個指針分配 memory ,如果要返回指向指針的指針,則需要相應地分配 memory :
Student **t = malloc(sizeof *t * n); // memory for the array of pointers
for(int i = 0; i < n; i++){ // memory for each individual pointer
t[i] = malloc(sizeof **t);
}
要稍后釋放指針,您還需要釋放先前分配的每個單獨的指針:
for(int i = 0; i < n; i++){ // memory for each individual pointer
free(students[i]);
}
free(students);
請注意,單個字符的說明符是%c
,需要更正 printf :
printf("Name: %s, Sex: %c, Age: %d\n", students[i]->name, students[i]->sex, students[i]->age);
// ^^
我要更改的另一件事是在strncpy
而不是 null 稍后終止字符串,我會讓 function 這樣做:
// one more byte and strncpy terminates the string for you
strncpy(t[i]->name, data[0], strlen(data[0]) + 1);
// ^^^^
// already corrected for the new pointer
更正了此處的問題是一種可能的替代方法,您可以使用sscanf
從一個 go 中的entry
解析結構中的所有元素,如果您願意:
Student **getstudents(int n)
{
Student **t = malloc(sizeof *t * n); // memory for the array of pointers
if (t == NULL)
{
perror("malloc");
exit(EXIT_FAILURE);
}
for (int i = 0; i < n; i++) // memory for each individual pointer
{
t[i] = malloc(sizeof **t);
if (t[i] == NULL)
{
perror("malloc");
exit(EXIT_FAILURE);
}
}
for (int i = 0; i < n; i++)
{
char entry[100];
if (fgets(entry, sizeof entry, stdin))
{
if (sscanf(entry, "%25s %c %d", t[i]->name, &t[i]->sex, &t[i]->age) != 3)
{
// deal with bad input
}
}
}
return t;
}
anastaciu 的回答指出了很多麻煩,但還有其他問題:
您的代碼中的另一個問題是您使用 '%s' 作為性別,因為性別是唯一的字符。 您應該使用 %c 否則 printf function 將嘗試解析字符串並獲得 SEGFAULT。
我也敦促您嚴格檢查每個 memory 分配。 總是。
我的pov修改后的代碼:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct student Student;
struct student{
char name[100];
int age;
char sex;
};
Student **getstudents(int);
void free_students(Student**, int);
int main(){
Student **students;
int n = 4;
students = getstudents(n);
for(int i = 0; i < n; i++){
if (students[i] != NULL) {
printf("Name: %s, Sex: %c, Age: %d\n", students[i]->name, students[i]->sex, students[i]->age);
}
}
free_students(students, n);
return 0;
}
Student **getstudents(int n){
Student **t = (Student **)malloc(sizeof(Student *)*n);
if (t==NULL) {
perror("Memory: can't allocate.");
return(NULL);
}
/* Input: <name> <sex> <age> */
char entry[100];
for(int i = 0; i < n; i++){
t[i]=NULL;
if (fgets(entry,100,stdin) != NULL) {
int readBytes = strlen(entry);
char newString[3][25];
int k,j,ctr;
j=0; ctr=0;
for(k=0;k<=readBytes;k++)
{
if(entry[k]==' '||entry[k]=='\0'||entry[k]=='\n')
{
newString[ctr][j]='\0';
ctr++;
j=0;
}
else
{
newString[ctr][j]=entry[k];
j++;
}
}
t[i] = (Student *)malloc(sizeof(Student));
if (t[i] == NULL) {
perror("Memory: can't allocate.");
return(NULL);
}
strncpy(t[i]->name, newString[0], strlen(newString[0]));
t[i]->name[strlen(newString[0])] = '\0';
t[i]->sex = *newString[1];
t[i]->age = atoi(newString[2]);
}
}
return t;
}
void free_students(Student **students, int n){
for(int i=0; i<n; i++){
free(students[i]);
}
free(students);
}
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