1. The Joy of
Programming
Understanding Pointers in C S.G. GaneSh
Pointers are the forte of C, are the most difficult to master, and programming with them is prone
to errors. But pointers are fun too! This month, we’ll look at some puzzles to understand some
interesting aspects of pointers in C.
In the following programs, assume that necessary header to decide, so check the answers first:
files are included. A1. This program will run fine without an assertion failure.
Q1. Will this program result in an assertion failure? Sizes of all pointer types are equal! This might be surprising
to many programmers, but it is easy to understand. Pointers
int main() { signify an address. In general, for a given implementation,
assert(sizeof(void *) == sizeof(int *)); the storage space required for storing an address is the same,
assert(sizeof(int *) == sizeof(int **)); irrespective of the type of pointer used.
} A2. This program results in a compiler error for the
expression ‘i + j’. Why? Pointers signify the address and it
Q2. What will this program print? is illogical to add two addresses. However, you can add an
integer value to an address; for example, an array-based
int main() { address is a pointer and to locate an array element, it is
int iarr[10]; enough to simply add an integer to that address. Pointer
int *i = &iarr[2], *j = &iarr[5]; subtraction is allowed; in this case, for example, ‘i – j’
int *k = i + j; indicates the number of array elements between them, which
int diff = j – i; is equivalent to the expression (&iarr[5] - & iarr[2]), and is
printf(“%d”, diff); always 3, irrespective of the size of int.
} A3. Old C compilers or modern C compilers in K&R C
mode (which refers to pre-ANSI C -- the original C language
Q3. Will this program work? defined by Dennis M. Ritchie) do not have strong type
checking and, hence, they will compile this fine.
int main() { In the underlying implementation, if the size of int and
int i = “C is often unpredictable!”; the size of the pointer are the same, then there is no problem
printf(i); in storing the address of the string literal in integer i. printf
} is a dumb routine and it will interpret the first argument as
a string (in fact, i has an address of a string literal). So, this
Q4. What does the following program print? program might compile and print: “C is often unpredictable!”
A4. Yes, this program will print “Joy”! Note that strncpy
int main() { returns a char* which is the address of the copied string.
char string[10]; Here, strncpy copies three characters and returns that string.
printf(strncpy(string ,”Joy of C”,3)[3] = ‘0’); Then, we do indexing on that returned char* and put the null
} terminator ‘0’ for that string in the index position [3]. The
printf gets “Joy” as the argument and prints it.
Q5. What does this following program print? A5) This program results in a compiler error for the
expression ‘&&i’. The ‘&&’ operator is a logical ‘and’ operator
int main() { and requires two operands. Ignoring this syntax issue, the
// assume that address of i is 0x1234ABCD more important problem is that the attempted expression
int i = 10; is illogical. It is possible to take ‘address of i’ with &i; but
int * ip = &i; address of ‘address of i’ cannot exist!
int **ipp = &&i;
By: S G Ganesh is a research engineer at Siemens
printf(“%x, %x, %x”, &i, ip, *ip);
(Corporate Technology), Bangalore. His latest book is ‘60
} Tips on Object Oriented Programming’ published by Tata
McGraw-Hill in December 2007. You can reach him at
Well, they don’t seem too difficult, do they? It is too soon sgganesh@gmail.com
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