bajishareef

Hi, you have called the function like mainmenu() without declaration. C treats that as int mainmenu(void); But in the function definition you written void mainmenu(void). Obviously this function wouldn't be called. SO declaration violation occured. Means the function which you have called has no declaration. As per the rules each …

you have declared as int a,b,c=1; You did not specify any storage class. By default all local variables of [B]auto[/B] type. The default value of auto type is garbage value. In for loop c*=b means c=c*b means you are multiplying the values of b and c. where as b is …

Its just the garbage value. a[3] means nothing.. and its value any thing. array bounded at a[2].

According to the standards of 'C' language there is nothing written about a function cannot return which has return type void. Any function which has void return type may not return value, but if we want we can return. Its not the compulsory condition. But we should not assign the …

May be you pressed a shortcut key for navigator. or any graphics oriented files not loaded successfully..

just put extra space infront of %c like scanf(" %c",&var);

can I use c++

pointers are supposed to handle already allocated memory. That can be either by user or predefined like VGA memory. But you assigned the address 0x2000. what's the meaning of that. That address belongs to which data type. By pointer arithmetic theory ++ means current address plus no. of bytes allocated …

arrow keys doesn't have the ascii codes. They have the scan codes. To obtain them we need to write like char ch; ch=getch(); ch=getch(); left arrow->75 right arrow->77 up arrow->72 down arrow->80 Happy.......

At run time just enter cntl+z which equals to end of file (EOF) character.

pointer variables are supposed to hold the address of compatible type of variables which are already declared. pointers supposed to indicate value at that address. You mentioned int *ptr=50. This is meaning less. first declare int type of variable and assign the value to it. Then pass the address of …

Hi I am new this blog. within if condition if you compare the variable with constant 'ma'. Its the semantic flaw. Out of range problem. If you compare with "m" or "ma" error will occur. Because double quoted characters are of string type and they cannot be compared with single …