我想将参数作为引用，以便在主函数中使用“nextfreeplace”。问题是，我并不真正理解将参数作为参考的术语。有人能帮忙吗。我也收到编译警告。
#include <stdio.h>
#include <stdlib.h>
/* these arrays are just used to give the parameters to 'insert',
to create the 'people' array */
char *names[7]= {"Simon", "Suzie", "Alfred", "Chip", "John", "Tim",
"Harriet"};
int ages[7]= {22, 24, 106, 6, 18, 32, 24};
/* declare your struct for a person here */
typedef struct{
char *names;
int ages;
} person;
static void insert (person **p, char *s, int n, int *nextfreeplace) {
*p = malloc(sizeof(person));
/*static int nextfreeplace = 0;*/
/* put name and age into the next free place in the array parameter here */
(*p)->names=s;
(*p)->ages=n;
/* make the parameter as reference*/
sscanf(nextfreeplace,"%d", *p);
/* modify nextfreeplace here */
(*nextfreeplace)++;
}
int main(int argc, char **argv) {
/* declare nextinsert */
int *nextfreeplace = 0;
/* declare the people array here */
person *p[7];
//insert the members and age into the unusage array.
for (int i=0; i < 7; i++) {
insert (&p[i], names[i], ages[i], nextfreeplace);
/* do not dereference the pointer */
}
/* print the people array here*/
for (int i=0; i < 7; i++) {
printf("The name is: %s, the age is:%i\n", p[i]->names, p[i]->ages);
}
/* This is the third loop for call free to release the memory allocated by malloc */
/* the free()function deallocate the space pointed by ptr. */
for(int i=0; i<7;i++){
free(p[i]);
}
}
最佳答案
这应该改为下面的代码，因为(*nextfreeplace)++;将尝试访问0x000000000处的地址，这可能会导致segmentation fault。
int nextfreeplace = 0;
/* declare the people array here */
person *p[7];
//insert the members and age into the unusage array.
for (int i=0; i < 7; i++) {
insert (&p[i], names[i], ages[i], &nextfreeplace);
/* do not dereference the pointer */
}
关于c - 将该参数作为引用，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/13009684/