问题
I came across this puzzle here. I can't figure out why NONE is not printed. Any ideas?
#include<stdio.h>
int main()
{
int a=10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}
回答1:
defa1ut: is a syntactically valid label, e.g. for a goto but not the default of the switch statement.
If you compile with gcc with enough warnings it will point this out:
ajw@rapunzel:/tmp > gcc -Wall -Wextra test.c
test.c: In function ‘main’: test.c:13:15: warning: label ‘defa1ut’ defined but not used
It's a good argument for building with warnings cranked up high and aiming for 0 warnings in every build.
回答2:
If defa1ut is a typo for default and the string "NONE" is printed:
This is because '1' and 1 is different.
'1' means the ASCII value of the character '1' whose value in decimal is 49. and 1 is an integer.
The first case will be true if the value of a is 49 or '1' , but as a=10 so it is neither equal to '1' nor equals to '2' and thus default is executed (if it existed, and defa1ut is not a typo).
If defa1ut is not a typo for default and simply nothing is printed:
In this case you have no default instead which look like it is defa1ut which will act as a normal label, so simply nothing will be printed.
回答3:
default is spelled wrong. and so that case is never reached. http://codepad.org/gQPA6p4s
#include<stdio.h>
int main()
{
int a=10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defalut:
printf("NONE\n");
mickey_mouse:
printf("No Mickey\n");
default :
printf("CORRECT DEFAULT\n");
}
return 0;
}
回答4:
Since defa1ut is not keyword, it should be addressed with a case statement.
回答5:
Why do you think it should be printed?
defa1ut is different from default
来源:https://stackoverflow.com/questions/7175348/need-help-solving-a-c-based-programming-puzzle