Q:What will be the output of the following code?
void main ()
{ int i = 0 , a[3] ;
a[i] = i++;
printf (“%d”,a[i]) ;
}
Ans: The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value.
Q:Why doesn’t the following code give the desired result?
A: int x = 3000, y = 2000 ;
long int z = x * y ;
Ans: Here the multiplication is carried out between two ints x and y, and the result that would overflow would be truncated before being assigned to the variable z of type long int. However, to get the correct output, we should use an explicit cast to force long arithmetic as shown below:
long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.
Q: Why doesn’t the following statement work?
A: char str[ ] = “Hello” ;
strcat ( str, ‘!’ ) ;
Ans: The string function strcat( ) concatenates strings and not a character. The basic difference between a string and a character is that a string is a collection of characters, represented by an array of characters whereas a character is a single character. To make the above statement work writes the statement as shown below:
strcat ( str, “!” ) ;
Q:How do I know how many elements an array can hold?
A: The amount of memory an array can consume depends on the data type of an array. In DOS environment, the amount of memory an array can consume depends on the current memory model (i.e. Tiny, Small, Large, Huge, etc.). In general an array cannot consume more than 64 kb. Consider following program, which shows the maximum number of elements an array of type int, float and char can have in case of Small memory model.
main( )
{
int i[32767] ;
float f[16383] ;
char s[65535] ;
}
Q:How do I write code that reads data at memory location specified by segment and offset?
A: Use peekb( ) function. This function returns byte(s) read from specific segment and offset locations in memory. The following program illustrates use of this function. In this program from VDU memory we have read characters and its attributes of the first row. The information stored in file is then further read and displayed using peek( ) function.
#include
#include
main( )
{
char far *scr = 0xB8000000 ;
FILE *fp ;
int offset ;
char ch ;
if ( ( fp = fopen ( “scr.dat”, “wb” ) ) == NULL )
{
printf ( “\nUnable to open file” ) ;
exit( ) ;
}
// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
fprintf ( fp, “%c”, peekb ( scr, offset ) ) ;
fclose ( fp ) ;
if ( ( fp = fopen ( “scr.dat”, “rb” ) ) == NULL )
{
printf ( “\nUnable to open file” ) ;
exit( ) ;
}
// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
{
fscanf ( fp, “%c”, &ch ) ;
printf ( “%c”, ch ) ;
}
fclose ( fp ) ;
}
Q: What is C language?
A: The C programming language is a standardized programming language developed in the early 1970s by Ken Thompson and Dennis Ritchie for use on the UNIX operating system. It has since spread to many other operating systems, and is one of the most widely used programming languages. C is prized for its efficiency, and is the most popular programming language for writing system software, though it is also used for writing applications.
Q: Inprintf() Function What is the output of printf(“%d”)?
A:1. When we write printf(“%d”,x); this means compiler will print the value of x. But as here, there is nothing after �%d� so compiler will show in output window garbage value.
2. When we use %d the compiler internally uses it to access theargument in the stack (argument stack). Ideally compiler determines the offset of the data variable depending on the format specification string. Now when we write printf(“%d”,a) then compiler first accesses the top most element in the argument stack of the printf which is %d and depending on the format string it calculated to offset to the actual datavariable in the memory which is to be printed. Now when only %d will be present in the printf then compiler will calculate the correct offset (which will be the offset to access the integer variable) but as the actual data object is to be printed is not present at that memory location so it will print what ever will be the contents of that memory location.
3. Some compilers check the format string and will generate an error without the proper number and type of arguments for things like printf(…) and scanf(…).
Q:malloc() Function- What is the difference between “calloc(…)” and “malloc(…)”?
A:1. calloc(…) allocates a block of memory for an array of elements of a certain size. By default the block is initialized to 0. The total number of memory allocated will be (number_of_elements * size).
malloc(…) takes in only a single argument which is the memory required in bytes. malloc(…) allocated bytes of memory and not blocks of memory like calloc(…).
2. malloc(…) allocates memory and returns a void pointer to the allocated space, or NULL if there is insufficient memory available.
calloc(…) allocates an array in memory with elements initialized to 0 and returns a pointer to the allocated space .
Q: In printf() Function- What is the difference between “printf(…)” and “sprintf(…)”?
A: sprintf(…) writes data to the character array whereas printf(…) writes data to the standard output device.
Q:Compilation How to reduce a final size of executable?
A:Size of the final executable can be reduced using dynamic linking for libraries.
Bro I have Confusion in question diff. b/w malloc n calloc--- in the ending of the 1st point u have written that malloc allocates bytes of memory not block of memory as calloc but in the next point u have written malloc allocates block of memory..
Thanks for correcting the error buddy.
Ya i wrote it wrong.
Actually in second line it was "malloc allocates bytes of memory and returns null"
1] a[i] = i++ is wrong for very different reasons.
According to you if a had been previously initialized this would have been correct ? no it isn't.
check out sequence points (http://en.wikipedia.org/wiki/Sequence_point) and this question (http://stackoverflow.com/questions/1969106/precendence-order-of-evaluation)
2] In general array cannot hold more than 64kb. That's a historical artifact. The 64kb limit on 8085 comes from 16 bit address lines, which are byte addressable, so total number of bytes = 2^16 = 65,536 or 64 kb.
(http://en.wikipedia.org/wiki/RAM_limit#16_address_bits.2C_16_address_pins)
3] When we use %d the compiler internally uses it to access the argument in the stack (argument stack).
This is according to the cdecl x86 convention but The standard (c89) offers no such guarantee. The compiler may as well pass the arguments through registers. see fastcall calling convention (http://stackoverflow.com/questions/672268/fastcall-gcc-example)
4]malloc(…) allocates memory and returns a void pointer to the allocated space, or NULL if there is insufficient memory available. calloc(…) allocates an array in memory with elements initialized to 0 and returns a pointer to the allocated space .
malloc() and calloc() both return pointer of type void *, which points to the first byte of allocated space or NULL in case of failure.
calloc could be naively implemented as:
void *calloc(size_t n, size_t size)
{
void *p = malloc(n * size);
memset(p, 0, n * size);
return p;
}
my mistake in naive calloc() implementation:
void *calloc(size_t n, size_t size)
{
void *p = malloc(n * size);
if (p)
memset(p, 0, n * size);
return p;
}