C语言 声明struct类型，将值插入该类型的数组中，并打印出数组
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Declare struct type, insert values into array of the type, and print out array
提问by Steven Liu
In this program, I would like to define a struct called person, and an insert function for inserting an element into unused space in the array which is declared as type person. Lastly, I would like to print out the results as standard output. Could anyone give me a hint for correct anything which is wrong? Cheers
在这个程序中，我想定义一个名为 person 的结构体，以及一个用于将元素插入到声明为 person 类型的数组中未使用空间的插入函数。最后，我想将结果打印为标准输出。任何人都可以给我一个提示以纠正任何错误吗？干杯
Error:
错误：
arrays.c:16:22: error: expected ')' before '[' token
arrays.c: In function 'main':
arrays.c:34:5: warning: implicit declaration of function 'insert'
arrays.c:41:5: warning: format '%s' expects type 'char *', but argument 2 has type 'char **'
Code
代码
#include <stdio.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 name;
int ages;
} person;
static void insert (p[], char *name, int ages) {
static int nextfreeplace = 0;
/* put name and age into the next free place in the array parameter here */
person p[0] = {&name, age};
/* modify nextfreeplace here */
nextfreeplace++;
}
int main(int argc, char **argv) {
/* 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]);
p[i]= p[i+1];
}
/* print the people array here*/
for (int i=0; i < 7; i++) {
printf("%s is %d years old\n", &names[i], ages[i]);
}
return 0;
}
回答by Tuntuni
The first problem is your struct person. You're declaring name as a charwhile it should be a char* (pointer)or a char[] (array).
第一个问题是你的结构人。您将 name 声明为char而它应该是char* (pointer)或char[] (array)。
typedef struct
{
char *name; //or char name[100];
int age;
}
person;
Next, your insertfunctionhas incorrect arguments. You don't want an array of persons (you could do it but this is simpler), you want a pointer to a person structso you can edit it.
接下来，您的插入函数的参数不正确。您不需要一组人员（您可以这样做，但这更简单），您需要一个指向人员结构的指针，以便您可以对其进行编辑。
static void insert(person *p, char *name, int age)
{
p->name = name;
p->age = age;
}
Finally, this is how you would populate your array and print it out:
最后，这是填充数组并将其打印出来的方式：
int main()
{
//names and ages...
person people[7];
for (int i = 0; i < 7; i++)
{
insert(&people[i], names[i], ages[i]);
}
for (int i = 0; i < 7; i++)
{
printf("name: %s, age: %i\n", people[i].name, people[i].age);
}
}
Example: http://ideone.com/dzGWId.
示例：http: //ideone.com/dzGWId。
回答by md5
You have a problem with the parameter p.
您的参数有问题p。
static void insert (p[], char *name, int ages)
You've forgotten its type (person). Then you redeclare it; the following instruction is invalid:
你忘记了它的类型 ( person)。然后你重新声明它；以下指令无效：
person p[0] = {&name, age};
In the function call, you don't use an array, but a case of an array. So your function should be:
在函数调用中，您不使用数组，而是使用数组的情况。所以你的功能应该是：
typedef struct
{
char *name;
int ages;
} person;
static void
insert (person *p, char *s, int n)
{
p->name = s;
p->ages = n;
}
And the call:
和电话：
insert (&p[i], names[i], ages[i]);
