我想开始一个Y和N问答程序。
#include <stdio.h>
#include <stdlib.h>
int main(){
char answer[256];
do {
print("\nDo you want to delete yourself of the record?\n");
scanf("%s", answer);
printf("%s", answer);
}while(answer != "Y" || answer != "N")
;
return 0;
}
如您所见,我声明了一个256个元素的char类型的变量,然后使用scanf记录了用户输入并将其存储在答案中。然后,只要用户输入大写的Y或N,循环就会一直询问。问题是,使用此实现,即使我输入Y或N,程序也会不断询问。是否应将char声明更改为单个字符?我已经尝试过了:
#include <stdio.h>
#include <stdlib.h>
int main(){
char answer;
do {
print("\nDo you want to delete yourself of the record?\n");
scanf("%c", answer);
printf("%c", answer);
}while(answer != 'Y' || answer != 'N')
;
return 0;
}
但我收到警告:
warning: format '%c' expects argument of type 'char *', but argument 2 has type int' [-Wformat=]
scanf("%c", answer);
有人对此问题有澄清吗?
此声明
然后,只要用户输入一个大写字母Y或N。
意味着用户将输入“ Y”或“ N”时,循环将停止其迭代,对吗?
此条件可以写为
strcmp( answer, "Y" ) == 0 || strcmp( answer, "N" ) == 0
因此,这种条件的否定(当循环将继续其迭代时,看起来像]
!( strcmp( answer, "Y" ) == 0 || strcmp( answer, "N" ) == 0 )
相当于
strcmp( answer, "Y" ) != 0 && strcmp( answer, "N" ) != 0
请注意,您必须比较字符串(使用C字符串函数strcmp
),而不是指向始终不相等的第一个字符的指针。
因此第一个程序的do-while循环中的条件应该是
do {
print("\nDo you want to delete yourself of the record?\n");
scanf("%s", answer);
printf("%s", answer);
}while( strcmp( answer, "Y" ) != 0 && strcmp( answer, "N" ) != 0 )
;
应该使用逻辑与运算符。
在第二个程序中,您必须像这样使用scanf的调用
scanf( " %c", &answer);
^^^^ ^
和相同的逻辑AND运算符
do {
print("\nDo you want to delete yourself of the record?\n");
scanf(" %c", &answer);
printf("%c", answer);
}while(answer != 'Y' && answer != 'N')
;