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C Questions-5
Posted Date: 05 Jun 2008 Resource Type: Articles/Knowledge Sharing Category: Placement Papers
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Posted By: ADITYA Member Level: Gold
Rating:  Points: 1
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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
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
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
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=f getch(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 inf ormation 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,
t i x
4 0 -4
3 1 -2
2 2 0
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;
{
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 f unction 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 f or
val2. This function returns the result of the operation performed by the
function 'func'. The function func has two integer parameters. The f ormal
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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Responses
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| Author: Lenin 05 Jun 2008 | Member Level: Diamond Points : 2 | Thanks for sharing these type of information. Surely the students of computer studies will use this.
The beauty is you have given the answer , also have explained the answers i mean with solutions, it is delightful
| | Author: Lenin 05 Jun 2008 | Member Level: Diamond Points : 2 | Thanks for sharing these type of information. Surely the students of computer studies will use this.
The beauty is you have given the answer , also have explained the answers i mean with solutions, it is delightful
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