Answer / susie

Answer :

1

Explanation

This is due to the close relation between the arrays and
pointers. N dimensional arrays are made up of (N-1)
dimensional arrays.

arr2D is made up of a 3 single arrays that contains 3
integers each .

The name arr2D refers to the beginning of all the 3 arrays.
*arr2D refers to the start of the first 1D array (of 3
integers) that is the same address as arr2D. So the
expression (arr2D == *arr2D) is true (1).

Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn’t change the value/meaning. Again arr2D[0] is the
another way of telling *(arr2D + 0). So the expression
(*(arr2D + 0) == arr2D[0]) is true (1).

Since both parts of the expression evaluates to true the
result is true(1) and the same is printed.

char inputString[100] = {0}; To get string input from the keyboard which one of the following is better? 1) gets(inputString) 2) fgets(inputString, sizeof(inputString), fp)

1 Answers  

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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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Predict the Output: int main() { int *p=(int *)2000; scanf("%d",2000); printf("%d",*p); return 0; } if input is 20 ,what will be print

2 Answers  

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How to use power function under linux environment.eg : for(i=1;i<=n;i++){ pow(-1,i-1)} since it alerts undefined reference to 'pow'.

2 Answers  

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3) Int Matrix of certain size was given, We had few valu= es in it like this. =97=97=97=97=97=97=97=97=97=97=97 1 = | 4 | | 5 | &= nbsp; | 45 =97=97=97=97=97=97=97=97=97=97=97 &n= bsp; | 3 | 3 | 5 | = | 4 =97=97=97=97=97=97=97=97=97=97=97 34 |&nbs= p; 3 | 3 | | 12 | &= nbsp; =97=97=97=97=97=97=97=97=97=97=97 3 | &nbs= p; | 3 | 4 | = | 3 =97=97=97=97=97=97=97=97=97=97=97 3 | = ; | | | = ; 3 | =97=97=97=97=97=97=97=97=97=97=97 &= nbsp; | | 4 | = ; | 4 | 3 We w= ere supposed to move back all the spaces in it at the end. Note: = If implemented this prog using recursion, would get higher preference.

0 Answers   RoboSoft,

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main() { int i=5,j=6,z; printf("%d",i+++j); }

2 Answers  

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Give a very good method to count the number of ones in a 32 bit number. (caution: looping through testing each bit is not a solution)

7 Answers   Microsoft,

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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++; } }

1 Answers  

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main() { char *p; p="%d\n"; p++; p++; printf(p-2,300); }

1 Answers  

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main(int argc, char **argv) { printf("enter the character"); getchar(); sum(argv[1],argv[2]); } sum(num1,num2) int num1,num2; { return num1+num2; }

1 Answers  

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main() { char * strA; char * strB = I am OK; memcpy( strA, strB, 6); } a. Runtime error. b. I am OK c. Compile error d. I am O

4 Answers   HCL,

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#include<stdio.h> #include<conio.h> void main() { int a=(1,2,3,(1,2,3,4); switch(a) { printf("ans:"); case 1: printf("1");break; case 2: printf("2");break; case 3: printf("1");break; case 4: printf("4");break; printf("end"); } getch(); }

0 Answers  

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