There are several misunderstandings in the posted code.
First there is a misunderstanding of char
versus string
. A char is for instance a single letter, a single special character like .
, ;
, etc. (see note1) while a string is a serie of chars. So
'y' is a char
"yes" is a string
You print "Enter a char" but from the code it's obvious that you really want "Enter a string".
This leads to the next problem. To input a string using scanf
you need to pass a "pointer to char". Your code pass "a pointer to pointer to char" due to the &
. Further the passed pointer must point to some memory. So you need:
char a[10]; // Make it an array of char so that it can hold a string
printf("Enter a string, max 9 characters");
scanf("%9s", a); // No & before a and width specifier used to avoid buffer overflow
Now this part
if (a = "yes")
is not the way to compare two strings in C. For that you need the function strcmp
- like:
if (strcmp(a, "yes") == 0)
Putting it together it's like:
int main()
{
char a[10];
printf("Enter a string, max 9 characters");
scanf("%9s", a);
if (strcmp(a, "yes") == 0){
printf("Number is 30");
}
else if (strcmp(a, "no") == 0)
{
printf("Number is 50");
}
else
{
printf("oops");
}
return 0;
}
That said, I don't understand why you print stuff like: "Number is 30" but that's kind of irrelevant here.
note1: The type char
is actually an integer type, i.e. a number, but the common use is to map these numbers to characters using ASCII encoding.