Answer / hasini

simple steps:


String s1 = "Hello, ";
String s2 = "world!";
String s3 = s1 + s2;
System.out.println(s3);
// prints out: Hello, world!

char *someFun1() { char temp[ ] = “string"; return temp; } char *someFun2() { char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’}; return temp; } int main() { puts(someFun1()); puts(someFun2()); }

2 Answers  

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void main() { int c; c=printf("Hello world"); printf("\n%d",c); }

2 Answers  

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int swap(int *a,int *b) { *a=*a+*b;*b=*a-*b;*a=*a-*b; } main() { int x=10,y=20; swap(&x,&y); printf("x= %d y = %d\n",x,y); }

1 Answers  

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Sorting entire link list using selection sort and insertion sort and calculating their time complexity

1 Answers   Infosys, Microsoft, NetApp,

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Write out a function that prints out all the permutations of a string. For example, abc would give you abc, acb, bac, bca, cab, cba. You can assume that all the characters will be unique.

5 Answers   IITR, Microsoft, Nike,

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There is a lucky draw held every day. if there is a winning number eg 1876,then all possible numbers like 1867,1687,1768 etc are the numbers that match irrespective of the position of the digit. Thus all these numbers qualify fr the lucky draw prize Assume there is no zero digit in any numbers. write a program to show all the possible winning numbers if a "winning number"is passed as an arguments to the function.

1 Answers   Nagarro,

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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); }

1 Answers  

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/*what is the output for*/ void main() { int r; printf("Naveen"); r=printf(); getch(); }

4 Answers  

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how can i search an element in an array

2 Answers   CTS, Microsoft, ViPrak,

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What is the output for the following program main() { int arr2D[3][3]; printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) ); }

1 Answers  

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how can u draw a rectangle in C

53 Answers   Accenture, CO, Codeblocks, Cognizant, HCL, Oracle, Punjab National Bank, SAP Labs, TCS, University, Wipro,

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void main() { char ch; for(ch=0;ch<=127;ch++) printf(“%c %d \n“, ch, ch); }

1 Answers  

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