C Aptitude Questions and answers with explanation

C Aptitude 30
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

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In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

–> Predict the output or error(s) for the following:

C aptitude 30.1

  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.

C aptitude 30.2

   main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}

Answer: g20fy

Explanation: Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C aptitude 30.3

     # 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.

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Write a C program to reverse a string using pointers.

C Strings:
Write a C program to reverse a string using pointers.
In this program, we reverse the given string by using the pointers. Here we use the two pointers to reverse the string, strptr holds the address of given string and in loop revptr holds the address of the reversed string.
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#include<stdio.h>
int main(){
int i=-1;
char str[100];
char rev[100];
char *strptr = str;
char *revptr = rev;
printf("Enter the string:n");
scanf("%s",str);
while(*strptr)
{
strptr++;
i++;
}
while(i >=0) {
strptr--;
*strptr = *revptr;
revptr++;
--i;
}
printf("nn Reversed string is:%s",rev);
return 0;
}



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K & R C Programs Exercise 7-4.

K and R C, Solution to Exercise 7-4:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
Write a private version of scanf analogous to minprintf from the previous section.
minscanf is similar to minprintf. This function collects characters from the format string until it finds an alphabetic character after a %. That is the localfmt passsed to scanf along with the appropriate pointer.
The arguments to scanf are pointers: a pointer to a format string and a pointer to the variable that receives the value from scanf.Read more about C Programming Language .
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#include <stdarg.h>
#include <stdio.h>
#include <ctype.h>
#define LOCALFMT 100

/* minscanf: minimal scanf with variable argument list */
void minscanf(char *fmt, ...)
{
va_list ap;
char *p, *sval;
char localfmt[LOCALFMT];
int i,c;
int *ival;
double *dval;
unsigned *uval;

va_start(ap, fmt); /* make ap point to the first unnamed arg */
for (p = fmt; *p; p++) {
if (*p != '%') {
localfmt[i++] = *p;
continue;
}
i = 0;
localfmt[i++] = '%';
while(*p(p+1) && !isalpha(*(p+1)))
localfmt[i++] = *++p;
localfmt[i++] = *(p+1);
localfmt[i] = '/0';
switch (*++p) {
case 'd':
case 'i':
ival = va_arg(ap, int *);
scanf(localfmt, ival);
break;

case 'u':

case 'o':

case 'x':

case 'X':


case 'f':
dval = va_arg(ap, double);
scanf(localfmt, dval);
break;

case 's':
sval = va_arg(ap, char *);
scanf(localfmt, sval);
break;
default:
scanf(localfmt);
break;
}
i = 0;
}
va_end(ap);
}


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K & R C Programs Exercise 5-20.

K and R C, Solution to Exercise 5-20:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
C Program to expand dcl to handle declarations with function argument types, qualifiers like const, and so on. Read more about C Programming Language .
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#include<stdio.h>
#include<string.h>
#include<ctype.h>
enum { NAME, PARENS, BRACKETS};
enum { NO, YES};
void dcl(void);
void dirdcl(void);
void errmsg(char *);
void dclspec(void);
int typespec(void);
int typeequal(void);
int compare(char **, char**);
void parmdcl(void);

int gettoken(void);
extern int tokentype;
extern char token[];
extern char name[];
extern char datatype[];
extern char out[];
extern int prevtoken;


//dcl:parse a declarator
void dcl(void)
{
int ns;
for(ns = 0; gettoken() == '*';)
ns++;
dirdcl();
while(ns --> 0)
strcat(out, "pointer to");
}


//dirdcl: parse a direct declaration
void dirdcl(void)
{
int type;
void parmdcl(void);
if(tokentype == '('){
dcl();
if(tokentype != ')')
errmsg("error:missimg)n");
}
else if(tokentype == NAME){
if(name[0] == '