[英]Using malloc with struct in C
So, this is my code: 所以,这是我的代码:
#include <stdio.h>
#include <string.h>
#include <assert.h>
#include <stdlib.h>
struct person{
char name[18];
int age;
float weight;
};
int main()
{
struct person *personPtr=NULL, person1, person2;
personPtr=(struct person*)malloc(sizeof*(struct person));
assert (personPtr!=NULL);
personPtr = &person1; // Referencing pointer to memory address of person1
strcpy (person2.name, "Antony"); //chose a name for the second person
person2.age=22; //The age of the second person
person2.weight=21.5; //The weight of the second person
printf("Enter a name: ");
personPtr.name=getchar(); //Here we chose a name for the first person
printf("Enter integer: ");
scanf("%d",&(*personPtr).age); //Here we chose the age of the first person
printf("Enter number: ");
scanf("%f",&(*personPtr).weight); //Here we chose the weithgt of the first person
printf("Displaying: "); //Display the list of persons
printf("\n %s, %d , %.2f ", (*personPtr).name, (*personPtr).age,(*personPtr).weight); //first person displayed
printf("\n %s, %d , %.2f",person2.name, person2.age, person2.weight); //second person displayed
free(personPtr);
return 0;
}
I get two errors and I don't know why. 我有两个错误,我不知道为什么。 Firstly, I don't think that I allocated the memory right, the first error is on the next line: 首先,我不认为我分配了内存,第一个错误在下一行:
personPtr=(struct person*)malloc(sizeof*(struct person));
It says that: 它说:
[Error] expected expression before ')' token [错误]')'标记之前的预期表达式
The second error that I get is on the line 我得到的第二个错误是在线上
personPtr.name=getchar();
Why cant I assign a name using getchar for a structure? 为什么不能使用getchar为结构分配名称? The error is: 错误是:
[Error] request for member 'name' in something not a structure or union [错误]要求成员“名称”使用非结构或联合的名称
sizeof*(struct person)
is a syntax error. sizeof*(struct person)
是语法错误。 It is seen by the compiler as an attempt to apply the sizeof
operator to *(struct person)
. 编译器将其视为尝试将sizeof
运算符应用于*(struct person)
。 Since you can't dereference a type, the compiler complains. 由于您不能取消引用类型,因此编译器会抱怨。 I think you meant to write the following: 我认为您打算编写以下内容:
personPtr = malloc(sizeof *personPtr);
It's the idiomatic way to allocate whatever personPtr
is pointing to. 这是分配personPtr
指向的内容的惯用方式。 Now the type is specified only where the pointer is defined, and that's a good thing. 现在,仅在定义指针的位置指定类型,这是一件好事。 You also don't need to cast the result of malloc
, since void*
is implicitly convertible to any pointer type. 您也不需要malloc
转换malloc
的结果,因为void*
可隐式转换为任何指针类型。
The second error is two-fold: 第二个错误是双重的:
name
is a fixed sized array. name
是固定大小的数组。 You cannot assign to an array using the assignment operator. 您不能使用赋值运算符分配给数组。 You can only assign to each individual element. 您只能分配给每个单独的元素。
getchar
returns a single character, not a string as you seem to expect. getchar
返回单个字符,而不是您期望的字符串。 To read a string, you can use scanf("%17s", personPtr->name)
. 要读取字符串,可以使用scanf("%17s", personPtr->name)
。 The 17 is the size of your buffer - 1, to protect against buffer overflow when scanf adds a NUL terminator to the string. 17是缓冲区的大小-1,以防止scanf在字符串中添加NUL终止符时防止缓冲区溢出。
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