main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Answer / susie
Answer :
Explanation:
Here we have only one pointer to type char and since we take
input in the same pointer thus we keep writing over in the
same location, each time shifting the pointer value by 1.
Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for
the first input suppose the pointer starts at location 100
then the input one is stored as
M
O
U
S
E
\0
When the second input is given the pointer is incremented as
j value becomes 1, so the input is filled in memory starting
from 101.
M
T
R
A
C
K
\0
The third input starts filling from the location 102
M
T
V
I
R
T
U
A
L
\0
This is the final value stored .
The first printf prints the values at the position q,
q+1 and q+2 = M T V
The second printf prints three strings starting from
locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
| Is This Answer Correct ? | 2 Yes | 0 No |
how to return a multiple value from a function?
find simple interest & compund interest
main() { int x=5; for(;x!=0;x--) { printf("x=%d\n", x--); } } a. 5, 4, 3, 2,1 b. 4, 3, 2, 1, 0 c. 5, 3, 1 d. none of the above
plz tell me the solution.......... in c language program guess any one number from 1 to 50 and tell that number within 8 asking question in yes or no...............
what is oop?
What is the output for the program given below typedef enum errorType{warning, error, exception,}error; main() { error g1; g1=1; printf("%d",g1); }
how can i search an element in an array
2 Answers CTS, Microsoft, ViPrak,
why nlogn is the lower limit of any sort algorithm?
#include<stdio.h> main() { const int i=4; float j; j = ++i; printf("%d %f", i,++j); }
create a C-code that will display the total fare of a passenger of a taxi if the driver press enter,the timer will stop. Every 10 counts is 2 pesos. Initial value is 25.00
Sir... please give some important coding questions asked by product companies..
main() { int i=5,j=6,z; printf("%d",i+++j); }