C program to sort a string

C Strings:
Write a C program to sort a string.
In this program we sort the string using bubble sort technique.
Bubble Sort is the simplest and easiest sorting technique. In this technique, the two successive items A[i] and A[i+1] are exchanged whenever A[i]>=A[i+1]. The larger values sink to the bottom of the array and hence it is called sinking sort. The end of each pass smaller values gradually “bubble” their way upward to the top(like air bubbles moving to surface of water) and hence called bubble sort.
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#include<stdio.h>
#include<string.h>
int main(){
    int i,j,n;
    char str[50],temp[50];
    printf("Enter a String:nn");
    scanf("%s",str);
    n=strlen(str);
    for(i=0;i<=n;i++)
    for(j=i+1;j<=n;j++){
        if(strcmp(str[i],str[j])>0){
            strcpy(temp,str[i]);
            strcpy(str[i],str[j]);
            strcpy(str[j],temp);
        }
    }
    printf("The sorted string is:%sn",str);
    return 0;
}
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C Aptitude Questions and answers with explanation

C Aptitude 21
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.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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.
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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 21.1

 char *Fun1()
{
char temp[ ] = “example";
return temp;
}
char *Fun2()
{
char temp[ ] = {‘e’, ‘x’,’a’,’m’,’p’,’l’,'e'};
return temp;
}
int main()
{
puts(Fun1());
puts(Fun2());
}


Answer: Garbage values.

Explanation: Both the functions suffer from the problem of dangling pointers. In Fun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function Fun2() also suffers from the same problem but the problem can be easily identified in this case.

C aptitude 21.2

   char *strexp()
{
char *temp = "example string";
return temp;
}
int main()
{
puts(strexp);
}

Answer: example string

Explanation: The character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

C aptitude 21.3

     main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}

Answer: -1

Explanation: Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i–!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

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C Aptitude Questions and answers with explanation

C Aptitude 20
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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Predict the output or error(s) for the following:

C aptitude 20.1

  main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}


Answer: 10

Explanation: The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives up to the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

C aptitude 20.2

   main()
{
int i=-1;
-i;
printf("i = %d, -i = %d n",i,-i);
}


Answer: i = -1, -i = 1

Explanation: -i is executed and this execution doesn’t affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

C aptitude 20.3

     main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}

Answer: Compiler error

Explanation: i is a constant. you cannot change the value of constant

C aptitude 20.4

      main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

Answer: garbagevalue..1

Explanation: p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer.

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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.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#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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C Aptitude Questions and answers with explanation

C Aptitude 19
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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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

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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 19.1

  main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}


Answer: 1==1 is TRUE

Explanation: When two strings are placed together (or separated by white-space) they are concatenated (this is called as “stringization” operation). So the string is as if it is given as “%d==1 is %s”. The conditional operator( ?: ) evaluates to “TRUE”.

C aptitude 19.2

  main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}



Answer: 2000 is a leap year

Explanation: An ordinary program to check if leap year or not.

C aptitude 19.3

     #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

Answer: Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation: arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error. Rule of Thumb: #defines are used for textual replacement whereas typedefs are used for declaring new types.

C aptitude 19.4

      int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

Answer: 30,20,10

Explanation: a'{‘ introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i’s value as 10.

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C Aptitude Questions and answers with explanation

C Aptitude 18
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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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 18.1

  main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}



Answer: 2 5 5

Explanation: In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the ‘’ termination character). The third sizeof is similar to the second one.

C aptitude 18.2

  main()
{
char not;
not=!2;
printf("%d",not);
}


Answer: 0
Explanation:! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

C aptitude 18.3

    #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}

Answer: TRUE

Explanation: The input program to the compiler after processing by the preprocessor is,
main()
{
  if(0)
    puts(“NULL”);
  else if(-1)
    puts(“TRUE”);
  else
    puts(“FALSE”);
 }
Preprocessor doesn’t replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence “TRUE” is printed.

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C Aptitude Questions and answers with explanation

C Aptitude 17
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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Predict the output or error(s) for the following:

C aptitude 17.1

  int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}


Answer: Runtime error: Abnormal program termination. assert failed (i<5), ,

Explanation: sserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use, #undef NDEBUG and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of.

C aptitude 17.2

  main()
{
int i=-1;
+i;
printf("i = %d, +i = %d n",i,+i);
}

Answer: i = -1, +i = -1

Explanation:Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

C aptitude 17.3

    main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}

Answer:Compiler error (at line number 4): size of v is Unknown.

Explanation: You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

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C Aptitude Questions and answers with explanation

C Aptitude 16
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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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 16.1

 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 16.2

  main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}

Answer: ck

Explanation:In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

C aptitude 16.3

    main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%sn”,x);
x++;
}
}

Answer:(blank space)
irl
rl
l

Explanation: Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘’) to the first location. Now the string becomes “irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

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C Aptitude Questions and answers with explanation

C Aptitude 15
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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Predict the output or error(s) for the following:

C aptitude 15.1

  main()
{
extern out;
printf("%d", out);
}
int out=100;


Answer: 100

Explanation: This is the correct way of writing the previous program 14.3

C aptitude 15.2

 main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

Answer: Compier error: Type mismatch in redeclaration of show.

Explanation:When the compiler sees the function show it doesn’t know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

C aptitude 15.3

    main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}

Answer:Compiler error: lvalue required..

Explanation: Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

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C Aptitude Questions and answers with explanation

C Aptitude 14
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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Predict the output or error(s) for the following:

C aptitude 14.1

  main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}


Answer: Compiler Error

Explanation: in the end of nested structure yy a member have to be declared

C aptitude 14.2

  main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

Answer: Linker error: undefined symbol ‘i’.

Explanation:extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn’t find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

C aptitude 14.3

     main()
{
printf("%d", out);
}

int out=100;

Answer: Compiler error: undefined symbol out in function main.

Explanation: The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main.

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