我有一个要工作的应用程序,在这里我需要使用结构类型的数组作为structure的成员。我已经创建了具有结构的模型,但是正如您所看到的,我需要一个作为接触结构内部数组的成员(动态地取决于用户的输入)。我正在通过从main传递一个名为contactPersonLength的变量来在getContacts函数中为其分配大小,但是一旦getContacts返回到contactPersonLength的主值时,变量就会更改,并且垃圾值会在main中打印。所以我在传递变量到打印功能时遇到了麻烦。请告诉我我错了,以及如何为结构成员的数组分配大小
struct address
{
char *name,*street,*district;
int doorNo;
};
struct contactPerson
{
char *name,*contactNumber;
};
struct contact
{
char *firstName,*lastName,*emailId;
struct address billingAddress;
struct address shippingAddress;
struct contactPerson contactPersons[];
};
void getContacts(struct contact *contacts,int n,int contactPersonLength)
{
int isSame,i,j;
for(i=0;i<n;i++)
{
printf(".......Enter Contact Details %d ........\n",i+1);
contacts[i].firstName = (char *)malloc(sizeof(char));
printf("Enter the First Name");
scanf("%s",contacts[i].firstName);
contacts[i].lastName = (char *)malloc(sizeof(char));
printf("Enter the Last Name");
scanf("%s",contacts[i].lastName);
contacts[i].emailId = (char *)malloc(sizeof(char));
printf("Enter the email id");
scanf("%s",contacts[i].emailId);
printf(".....Billing address Details.....\n");
contacts[i].billingAddress.name = (char *)malloc(sizeof(char));
printf("Enter the name");
scanf("%s",contacts[i].billingAddress.name);
printf("Enter the DoorNo");
scanf("%d",&contacts[i].billingAddress.doorNo);
contacts[i].billingAddress.street = (char *)malloc(sizeof(char));
printf("Enter the Street name");
scanf("%s",contacts[i].billingAddress.street);
contacts[i].billingAddress.district = (char *)malloc(sizeof(char));
printf("Enter the District");
scanf("%s",contacts[i].billingAddress.district);
printf(".....Shipping Address Details....\n");
printf("Is your Shipping Address same as Billing Address Press 1 Or else Press 0");
scanf("%d",&isSame);
if(isSame==1)
{
contacts[i].shippingAddress = contacts[i].billingAddress;
}
else
{
contacts[i].shippingAddress.name = (char *)malloc(sizeof(char));
printf("Enter the name");
scanf("%s",contacts[i].shippingAddress.name);
printf("Enter the DoorNo");
scanf("%d",&contacts[i].shippingAddress.doorNo);
contacts[i].shippingAddress.street = (char *)malloc(sizeof(char));
printf("Enter the Street name");
scanf("%s",contacts[i].shippingAddress.street);
contacts[i].shippingAddress.district = (char *)malloc(sizeof(char));
printf("Enter the District");
scanf("%s",contacts[i].shippingAddress.district);
}
printf(" ContactPersonLength %d \n",contactPersonLength);
contacts[i].contactPersons[contactPersonLength];
for(j=0;j<contactPersonLength;j++)
{
printf(".....ContactPerson %d.....\n",j+1);
contacts[i].contactPersons[j].name = (char *)malloc(sizeof(char));
printf("Enter Contact Person Name");
scanf("%s",contacts[i].contactPersons[j].name);
contacts[i].contactPersons[j].contactNumber = (char *)malloc(sizeof(char));
printf("Enter Contact Person Contact Number");
scanf("%s",contacts[i].contactPersons[j].contactNumber);
}
}
}
void main()
{
struct contact contacts[n];
getContacts(contacts,n,contactPersonLen);
}
答案 0 :(得分:3)
我建议您使用以下功能:
char buffer[256];
for(i=0;i<n;i++)
{
printf(".......Enter Contact Details %d ........\n",i+1);
do {
printf("Enter the First Name");
} while (scanf("%255s",buffer)!=1);
contacts[i].firstName = malloc(strlen(buffer)+1);
strcpy(contacts[i].firstName, buffer);
//...etc
因此,您使用单个缓冲区来获取输入,然后完全按照输入的大小分配内存,再加上一个终止的空字符,然后将缓冲区复制到该内存并将缓冲区重新用于下一个输入。
注意:char的大小始终为1,因此不需要sizeof(char)
。另外,不应强制转换malloc
的结果。
您在do...while
循环中询问输入,并检查scanf
是否能够读取输入。在scanf中,格式说明符将输入限制为255个字符,因此缓冲区不会溢出。
答案 1 :(得分:0)
(char *)malloc(sizeof(char));
仅分配一个字节的内存。如果要存储一个字符串,例如20个字符长,则代码为:
(char *)malloc(sizeof(char) * 20 + 1); //explicitly showed +1 for clarity, based on comment.
编辑
通常,最好在malloc
中设置数组的最大可能大小。但是,如果大小太大,和/或您担心内存,请执行以下操作:
contacts[i].firstName = (char *)malloc(sizeof(char) * MAX_POSSIBLE_SIZE);
printf("Enter the First Name");
scanf("%s",contacts[i].firstName);
char *temp = (char *)realloc(contacts[i].firstName, strlen(contacts[i].firstName) + 1);
if(temp!=NULL) contacts[i].firstName = temp;
对每个字符串执行此操作。