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

struct Foo { char *pName; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); strcpy(obj->pName,"Your Name"); printf("%s", obj->pName); } a. Your Name b. compile error c. Name d. Runtime error

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how to programme using switch statements and fuctions, a programme that will output two even numbers, two odd numbers and two prime numbers of the users chioce.

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posted by surbhi just now main() { float a = 5.375; char *p; int i; p=(char*)&a; for(i=0;i<=3;i++) printf("%02x",(unsigned char) p[i]); } how is the output of this program is :: 0000ac40 please let me know y this output has come

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What is the hidden bug with the following statement? assert(val++ != 0);

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Design an implement of the inputs functions for event mode

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Write a program to receive an integer and find its octal equivalent?

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