Interested to Buy Any Domain ? << Click Here >> for more details...
main()
{
int i = -3,j=2,k=0,m;
m= ++i || ++j && ++k;
printf("%d%d%d",i,j,k,m);
}
- 7 Answers
- 25984 Views
- HCL, I also Faced
- E-Mail Answers
Answers were Sorted based on User's Feedback
Answer / harend
results :
i=-2
j=2
k=0
m=1
First, the '&&' part is to be considered over '||'.
AS follow: m = ++i||++j&&++k (is given)
what ever be the result of (++j&&++k),the value of m =1
,since the new value i= -2 (i.e a non zero value so taken as
true or 1)
so,
1||(++j&&++k) will always be true, that is 1 . compiler
ignores ++j ,++k and only consider ++i.
thank you !
| Is This Answer Correct ? | 25 Yes | 6 No |
Answer / ep
After the execution of the m evaluation the variables can be:
i = -2
j = 2
k = 0
m = 1
All of of this is because compilers do NOT completely
evalute expressions if they can short cut the evaluation.
Anyway, this is very bad programming.
| Is This Answer Correct ? | 18 Yes | 10 No |
Answer / mayur dharmik
printf("%d%d%d",i,j,k,m);
it has only 3 %d.
So, it will print only 3 value.
| Is This Answer Correct ? | 9 Yes | 2 No |
Answer / mayur dharmik
output
-220
i.e,
i=-2, j=2, k=0.
it will print only 1st three value.
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / nidhi yadav
here logical && has higher priority than ||. so the block
(++j&&++k)will execute first and ans of this will be (3&&1=0)
since value increment first due to preincrement operator. now
the block(++i||0)will execute as (-2||0=1) since || opertor always gives true value except(0||0=0).
thus ans will be i=-2,j=3,k=1,m=1
| Is This Answer Correct ? | 2 Yes | 7 No |
Explain the concept of "dangling pointers" in C.
write a program to swap two numbers without using temporary variable?
write a program to convert a expression in polish notation (postfix) to inline (normal) something like make 723+* (2+3) x 7 (not sure) just check out its mainly printing expression in postfix form to infix
The OS is a program that uses various data structures. Like all programs in execution, you can determine the performance and other behavior of the OS by inspecting its state - the values stored in its data structures. In this part of the assignment, we study some aspects of the organization and behavior of a Linux system by observing values of kernel data structures exposed through the /proc virtual file system. The /proc virtual file system: Linux uses the /proc file system to collect information from kernel data structures. The /proc implementation provided with Linux can read many different kernel data structures. If you cd to /proc on a Linux machine, you will see a number of files and directories at that location. Files in this directory subtree each corresponds to some kernel data structure. The subdirectories with numeric names contain virtual files with information about the process whose process ID is the same as the directory name. Files in /proc can be read like ordinary ASCII files. You can open each file and read it using library routines such as fgets() or fscanf(). The proc (5) manual page explains the virtual files and their content available through the /proc file system. Requirements in detail: In this part, you are asked to write a program to report the behavior of the Linux kernel. Your program should run in two different versions. The default version should print the following values on stdout: • Processor type • Kernel version • The amount of memory configured into this computer • Amount of time since the system was last booted A second version of the program should run continuously and print lists of the following dynamic values (each value in the lists is the average over a specified interval): • The percentage of time the processor(s) spend in user mode, system mode, and the percentage of time the processor(s) are idle • The amount and percentage of available (or free) memory • The rate (number of sectors per second) of disk read/write in the system • The rate (number per second) of context switches in the kernel • The rate (number per second) of process creations in the system If your program (compiled executable) is called proc_parse, running it without any parameter should print out information required for the first version. Running it with two parameters "proc_parse <read_rate> <printout_rate>" should print out information required for the second version. read_rate represents the time interval between two consecutive reads on the /proc file system. printout_rate indicates the time interval over which the average values should be calculated. Both read_rate and printout_rate are in seconds. For instance, proc_parse 2 60 should read kernel data structures once every two seconds. It should then print out averaged kernel statistics once a minute (average of 30 samples). The second version of your program doesn't need to terminate.
Explain what is a 'null pointer assignment' error? Explain what are bus errors, memory faults, and core dumps?
main() { int i=5; printf("%d%d%d%d",i++,i--,i); }
Is there any demerits of using pointer?
where are auto variables stored? What are the characteristics of an auto variable?
who is first prime minister in india??
Disadvantages of C language.
What is int main () in c?
Combinations of fibanocci prime series
C 
