Answer / anil kumar

Static variables are used for internal contextual
communication purpose.
non static variables are not used for contextual
communication
for that please go through the below code:

static int i=10;
int main()
{
int x=20;
Printf(“%d %d”,x, i);
Fun();
return 0;
}
Void Fun()
{
Printf(“%d”, i);
}

In the above code “i” is the static variable and “x “is the
local variable

Answer / parth ujenia

main()
{
int i=5;

while(i!=0)
{
printf("%d",i--);
main();
}
getch();
}

output: 54321

Answer / vignesh1988i

when we have declared a variable as static...... we cant
initilize it again....... the meaning of static storage
class is for only one time initilization.... whenever the
compailer come accross the same static keyword ,... the
present value in that variable will get printed as the
compailer ignores the line static.........
eg:::::
#include<stdio.h>
#include<conio.h>
void main()
{
for(int i=0;i<3;i++)
{
static int p=1;
printf("%d ",p);
p++;
}
output will be:: 1 2 3
since the compailer ignores the static int p=1 after it
initilizs once...... and one more thing.. when we refer
variable p after the loop structure it will give an error
that::: "UNDEFINED SYMBOL 'P'" ,because the scope of this
static is only under the block and not ourtside.....

non static :: it is called as automatic storage class.... in
programs we would have given as;;
int a; or char sd; etc...
these inside the compailer treated as automatic storage
class..... the scope of this storage class is only undere
the block... after comming out it dies......

eg:::::::
#include<stdio.h>
#include<conio.h>
void main()
{
int a=12;
{
a=90;
printf("%d",a);
}
printf("%d",a);
}
the output will be::::::: 90 12.... because of the above
mentioned scope.....

i think you can clearely understand the concept....... if
you didnt understand ... send mail to me.. we can
discuss.... softvig_88@yahoo.com...
thank you

Write a C/C++ program that connects to a MySQL server and checks if the InnoDB plug-in is installed on it. If so, your program should print the total number of disk writes by MySQL

0 Answers  

---

Explain what are the standard predefined macros?

0 Answers  

---

the constant value in the case label is followed by a a) semicolon b) colon c) braces d) none of the above

0 Answers  

---

How do I create a directory? How do I remove a directory (and its contents)?

0 Answers  

---

Explain how do I determine whether a character is numeric, alphabetic, and so on?

0 Answers  

---

You have an int array with n elements and a structure with three int members. ie struct No { unsigned int no1; unsigned int no2; unsigned int no3; }; Point1.Lets say 1 byte in the array element is represented like this - 1st 3 bits from LSB is one number, next 2 bits are 2nd no and last 3 bits are 3rd no. Now write a function, struct No* ExtractNos(unsigned int *, int count) which extracts each byte from array and converts LSByte in the order mentioned in point1.and save it the structure no1, no2, no3. in the function struct No* ExtractNos(unsigned int *, int count), first parameter points to the base address of array and second parameter says the no of elements in the array. For example: if your array LSB is Hex F7 then result no1 = 7, no2 = 2, no3 = 7. In the same way convert all the elements from the array and save the result in array of structure.

2 Answers   Qualcomm,

---

what are the advantages & disadvantages of unions?

2 Answers  

---

How can you access memory located at a certain address?

0 Answers  

---

What will be printed as the result of the operation below: #include<..> int x; int modifyvalue() { return(x+=10); } int changevalue(int x) { return(x+=1); } void main() { int x=10; x++; changevalue(x); x++; modifyvalue(); printf("First output:%d\n",x); x++; changevalue(x); printf("Second output:%d\n",x); modifyvalue(); printf("Third output:%d\n",x); }

2 Answers  

---

Are negative numbers true in c?

0 Answers  

---

What is the purpose of sprintf?

0 Answers  

---

Can you mix old-style and new-style function syntax?

0 Answers  

---