- main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
- What are the following notations of defining functions known as?
i. int abc(int a,float b) {
/* some code */
}
ii. int abc(a,b)
int a; float b; {
/*
some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche
notation
- main(){
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to %
since it is incremented twice and again decremented by 2, it points to '%d\n'
and 300 is printed.
- main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(;
}
abc(char a[]){
a++;
printf("%c",*;
a++;
printf("%c",*;
}
Explanation:
The base address is modified
only in function and as a result a points to 'b' then after incrementing to 'c'
so bc will be printed.
- func(a,b)
int a,b;{
return( a= (a==b) );
}
main(){
int process(),func();
printf("The value of
process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3
parameters - 1, a pointer to another function
2 and 3, integers. When this function is invoked from main, the
following substitutions for formal parameters take place: func for pf, 3 for
val1 and 6 for val2. This function returns the result of the operation
performed by the function 'func'. The function func has two integer parameters.
The formal parameters are substituted as 3 for a and 6 for b. since 3 is not
equal to 6, a==b returns 0. therefore the function returns 0 which in turn is
returned by the function 'process'.
- void main(){
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I"
is declared as static, hence memory for I will be allocated for only once, as
it encounters the statement. The function main() will be called recursively
unless I becomes equal to 0, and since main() is recursively called, so the
value of static I ie., 0 will be printed every time the control is returned.
- void main(){
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret){
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie.,
the function name and the argument name can be the same.
Firstly, the function ret() is called in which the
sizeof(float) ie., 4 is passed, after
the first expression the value in ret will be 6, as ret is integer hence the
value stored in ret will have implicit type conversion from float to int. The
ret is returned in main() it is printed after and preincrement.
- void main(){
char a[]="12345\0";
int i=strlen(;
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose
length will be counted from 0 till the null character. Hence the 'I' will hold
the value equal to 5, after the pre-increment in the printf statement, the 6
will be printed.
- void main(){
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
- void main(){
int i;
char a[]="\0";
if(printf("%s\n",)
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many
characters does it print. Hence printing a null character returns 1 which makes
the if statement true, thus "Ok here" is printed.
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