- What is the output for the program given below?
typedef enum errorType{warning, error, exception,}error;
main() {
error g1;
g1=1;
printf("%d",g1);
}
Answer:
Compiler error: Multiple declaration for error
Explanation:
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note:
The extra comma in the declaration,enum errorType{warning, error, exception,} is not an error. An extra comma is valid and is provided just for programmer’s convenience.
- typedef struct error{int warning, error, exception;}error;
main(){
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer:
1
Explanation:
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
Note:
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!
- #ifdef something
int some=0;
#endif
main(){
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
- #if something == 0
int some=0;
#endif
main(){
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
0 0
Explanation:
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.
- What is the output for the following program?
main(){
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer:
1
Explanation:
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays. arr2D is made up of a 3 single arrays that contains 3 integers each .
arr2D
arr2D[1]
| arr2D[2] | |
| arr2D[3] |
The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.
- void main(){
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer:
You can answer this if you know how values are represented in memory
Explanation:
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.
- int swap(int *a,int *b){
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main() {
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer:
x = 20 y = 10
Explanation:
This is one way of swapping two values. Simple checking will help understand this.
- main(){
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
- main() {
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
- main(){
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.
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