some c apps
1) main()
{
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with
a signed value. Since the both types doesn't match,
signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition
becomes false and control comes out of the loop.
2) 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.
3 ) 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
4) 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.
5) 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.
6 ) 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'.
7) 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.
8) 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.
9 ) 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.
10) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
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