Answer / hello

int main(int argc, char * argv[])
{
//argc has count of no of arguments on command line.
// argv[] contains array of strings seperated by space.
// so argv[1] will give u, second argument, with first
// argument being ur ./a.out(in unix)/test.exe(in win)
// so this way u can access CL args.

// ur code.
}

Answer / rakesh

int main(int argc,char *argv[])
{
int i;
for(i=0;i<argc;i++)
cout<<argv[i];
}

Answer / pb

argc is nothing but Argument Count
argv is nothing Argument Vector (or) Argument Value which
is array of Strings.

The first argument to the Command line is always the
Program Name.

argc contains no. of Arguments that it passed including the
Program Name.So Leave the 1st Argument,

int i;
for(i=1;i<=argc;i++)
cout<<argv[i];

Answer / padmaraj

By using -->
int main(int argc, char * argv[])

In this statement the variable argc holds the at least one
argument of type int , the name of the file it self first
argument and *argv[] is array of char which takes the name
of file.......
If any correction plz E-mail me ...
Bye

main() { int i=4,j=7; j = j || i++ && printf("YOU CAN"); printf("%d %d", i, j); }

1 Answers  

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main() { char not; not=!2; printf("%d",not); }

1 Answers  

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main() { clrscr(); } clrscr();

2 Answers  

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#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d..%d",*p,*q); }

1 Answers  

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Is the following code legal? typedef struct a { int x; aType *b; }aType

1 Answers  

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main() { char *a = "Hello "; char *b = "World"; clrscr(); printf("%s", strcpy(a,b)); } a. “Hello” b. “Hello World” c. “HelloWorld” d. None of the above

4 Answers   Corporate Society, HCL,

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Question: We would like to design and implement a programming solution to the reader-writer problem using semaphores in C language under UNIX. We assume that we have three readers and two writers processes that would run concurrently. A writer is to update (write) into one memory location (let’s say a variable of type integer named temp initialized to 0). In the other hand, a reader is to read the content of temp and display its content on the screen in a formatted output. One writer can access the shared data exclusively without the presence of other writer or any reader, whereas, a reader may access the shared memory for reading with the presence of other readers (but not writers).

1 Answers  

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union u { struct st { int i : 4; int j : 4; int k : 4; int l; }st; int i; }u; main() { u.i = 100; printf("%d, %d, %d",u.i, u.st.i, u.st.l); } a. 4, 4, 0 b. 0, 0, 0 c. 100, 4, 0 d. 40, 4, 0

1 Answers   HCL,

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Develop a routine to reflect an object about an arbitrarily selected plane

0 Answers  

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#define a 10 void foo() { #undef a #define a 50 } int main() { printf("%d..",a); foo(); printf("%d..",a); return 0; } explain the answer

1 Answers  

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Is this code legal? int *ptr; ptr = (int *) 0x400;

1 Answers  

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what will be the output of this program? void main() { int a[]={5,10,15}; int i=0,num; num=a[++i] + ++i +(++i); printf("%d",num); }

3 Answers   Wipro,

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