91) In
the following pgm add a stmt in the
function fun such that the address of
'a' gets stored in
'j'.
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.
92) 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
93) 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.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
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.
95) 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'.
96) 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.
97) 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.
98) void
main()
{
char a[]="12345\0";
int i=strlen(a);
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.
99) void
main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void
main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
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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