81) main(int
argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2) int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the
function sum without converting it to integer values.
82) # include
<stdio.h> int one_d[]={1,2,3}; main()
{
int *ptr; ptr=one_d; ptr+=3; printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
83) #
include<stdio.h>
aaa() {
printf("hi");
} bbb(){ printf("hello");
} ccc(){ printf("bye");
}
main()
{
int (*ptr[3])(); ptr[0]=aaa; ptr[1]=bbb; ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type
int.ptr[0] is assigned to address
of the function aaa.
Similarly ptr[1] and
ptr[2] for bbb
and ccc respectively. ptr[2]() is
in effect of writing ccc(), since ptr[2] points to ccc.
85) #include<stdio.h>
main()
{
FILE *ptr; char i;
ptr=fopen("zzz.c","r"); while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked
against NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to
determine the truth value of the boolean expression. So the statement following
the if statement is not executed. The values of i and j remain unchanged and
get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the
information from the accumulator. Here _AH is the pseudo global variable
denoting the accumulator. Hence, the value of the accumulator is set 1000 so
the function returns value 1000.
88) int i;
main(){ int t;
for (
t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the
values during execution will be,
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right.
Only the rightmost value is returned and the other values are evaluated and
ignored. Thus the value of last variable y is returned to check in if. Since it
is a non zero value if becomes true so, "hello" will be printed.
90) 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.
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;
{
}
main()
{
return( a= (a==b) );
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);
}
same.
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the
argument name can be the
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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