List of All Companies that hire Freshers Off Campus 📈
1.Adobe: www.adobe.com/careers.html
2.Accolite: www.accolite.com/careers/
3.Airtel: www.airtel.in/careers/
4.Amazon: www.amazon.jobs/
5.Cadence Design Systems: www.cadence.com/en_US/company/careers.html
6.CleverTap: clevertap.com/careers/
7.Cisco: jobs.cisco.com/
8.Capgemini: www.capgemini.com/careers/
9.Cognizant: careers.cognizant.com/global/en
10.Zoho: www.zoho.com/careers/
1.Adobe: www.adobe.com/careers.html
2.Accolite: www.accolite.com/careers/
3.Airtel: www.airtel.in/careers/
4.Amazon: www.amazon.jobs/
5.Cadence Design Systems: www.cadence.com/en_US/company/careers.html
6.CleverTap: clevertap.com/careers/
7.Cisco: jobs.cisco.com/
8.Capgemini: www.capgemini.com/careers/
9.Cognizant: careers.cognizant.com/global/en
10.Zoho: www.zoho.com/careers/
Adobe
Let’s Adobe together. | Adobe Careers
At Adobe, we believe that great ideas come from everywhere in the company, and the next one could be yours. Bring your talents to Adobe and help us create the future.
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Here is website link : https://www.hackerrank.com/skills-verification
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Join our WhatsApp channel or group for more updates
https://chat.whatsapp.com/G6q3SsT2tXBFiSEvSwSjIj
Link is in bio 📍❤️
HackerRank
Join over 23 million developers in solving code challenges on HackerRank, one of the best ways to prepare for programming interviews.
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Char ch = 21;
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
Anonymous Quiz
27%
34
24%
-88
14%
-68
35%
18
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What is the output of this program?
#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}
#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}
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Untold Coding
Char ch = 21;
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
In this question we use formal a×2^k and
-(2^n -k) here k is the number and n is the no. of bit
Therefore
We have n = 8 , (char size)
K = 21
- ( 2⁸ -168) = - (256-168)
= -88
-(2^n -k) here k is the number and n is the no. of bit
Therefore
We have n = 8 , (char size)
K = 21
- ( 2⁸ -168) = - (256-168)
= -88
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Untold Coding
#include <stdio.h> int main() { int i = -1; for (++i; i++; i++) printf("Untoldcoding"); return 0; }
Think and give answer ... In quiz
Untold Coding
#include <stdio.h> int main() { int i = -1; for (++i; i++; i++) printf("Untoldcoding"); return 0; }
Explanation
1. Initialization : The loop starts with
++i
, which increments i
from -1 to 0.2. Condition : The condition
i++
is checked next. Since i++
returns the value of i
before incrementing, the loop checks if 0 is true, which it is not (0 is considered false in C).3. Increment: If the condition were true, the loop would proceed to increment
i
with i++
again, but this part doesn't get executed in this case because the condition i++
evaluates to false on the first check.Due to the condition
i++
being false when i
is 0, the loop doesn't execute even once. Therefore, "Untoldcoding" is never printed.Still have any doubt ? Drop a comment 👇
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Consider the following C program:
#include <stdio.h>
int main( )
{
int i, j, k = 0;
j=2*3/4+2.0/5+8/5;
k -= --j;
for (i=0; i<5; i++)
{
switch(i+k)
{
case 1:
case 2: printf("\n%d", i+k);
case 3: printf("\n%d", i+k)
default: printf("\n%d", i+k);
}
}
return 0;
}
#include <stdio.h>
int main( )
{
int i, j, k = 0;
j=2*3/4+2.0/5+8/5;
k -= --j;
for (i=0; i<5; i++)
{
switch(i+k)
{
case 1:
case 2: printf("\n%d", i+k);
case 3: printf("\n%d", i+k)
default: printf("\n%d", i+k);
}
}
return 0;
}
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Untold Coding
Consider the following C program: #include <stdio.h> int main( ) { int i, j, k = 0; j=2*3/4+2.0/5+8/5; k -= --j; for (i=0; i<5; i++) { switch(i+k) { case 1: case 2: printf("\n%d", i+k); …
Explanation
Total printf Calls
Summing up all the outputs from each iteration:
Iteration 1: 1 time
Iteration 2: 1 time
Iteration 3: 3 times
Iteration 4: 3 times
Iteration 5: 2 times
Total number of printf calls: 1 + 1 + 3 + 3 + 2 = 10
Still have any confusion, drop a comment 👇
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#include <stdio.h>
int main()
{
int i;
for(i=1; i<=10; i++){
if(i>5){
break;
}
}
printf("%d",i);
return 0;
}
int main()
{
int i;
for(i=1; i<=10; i++){
if(i>5){
break;
}
}
printf("%d",i);
return 0;
}
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