我正在尝试创建一个简单的程序,将十个“宠物”存储到一个数组中。每个结构包含必须通过函数访问的数据。由于某些原因,这似乎不像我所期望的那样。有谁知道为什么程序提示输入名称,然后在没有再次提示用户的情况下运行程序的其余部分?
#include<stdlib.h>
#include<stdio.h>
#include <string.h>
struct Pet {
char name[50]; //name
char type[50]; //type
char owner[50]; //owner
};
void setPetName(struct Pet *pet, char *name){
memcpy(pet->name,name, 50);
}
void setPetType(struct Pet *pet, char *type){
memcpy(pet->type,type, 50);
}
void setOwner(struct Pet *pet, char *owner){
memcpy(pet->owner,owner, 50);
}
char* getName(struct Pet *pet){
return pet->name;
}
char* getType(struct Pet *pet){
return pet->type;
}
char* getOwner(struct Pet *pet){
return pet->owner;
}
void printPetInfo(struct Pet *pet){
printf("Pet's name is %s, Pet's type is %s, Pet's owner is %s", pet->name, pet->type, pet->owner);
}
int main(){
struct Pet Pets[9];
int index;
char name[50], type[50], owner[50];
for (index=0; index<9; index++){
struct Pet pet;
printf("Please enter pet's name ");
scanf("%s\n", name);
setPetName(&pet, name);
printf("Please enter pet's type ");
scanf("%s\n", type);
setPetType(&pet, type);
printf("Please enter pet's owner ");
scanf("%s\n", owner);
setOwner(&pet, owner);
printPetInfo(&pet);
Pets[index]=pet;
}
return 0;
}发布于 2014-10-27 00:59:37
首先,不能在字符中保持字符串:
char name, type, owner;相反,您需要一个char数组(例如,char name[50]; )
然后扫描字符串的格式是%s,而不是&s。
scanf("&s\n", name);最后,如果要打印字符串,请使用格式%s,而不是%c (%c是打印单个字符)。
https://stackoverflow.com/questions/26579759
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