C Puzzles : Questions and Answers - Level 2
Questions
#includemain() { int *a, *s, i; s = a = (int *) malloc( 4 * sizeof(int)); for (i=0; i<4; i++) *(a+i) = i * 10; printf("%d\n", *s++); printf("%d\n", (*s)++); printf("%d\n", *s); printf("%d\n", *++s); printf("%d\n", ++*s); }
#includevoid fn(int); static int val = 5; main() { while (val --) fn(val); printf("%d\n", val); } void fn(int val) { static int val = 0; for (; val < 5; val ++) printf("%d\n", val); }
#includemain() { typedef union { int a; char b[10]; float c; } Union; Union x, y = { 100 }; x.a = 50; strcpy (x.b, "hello"); x.c = 21.50; printf ("Union 2 : %d %s %f\n", x.a, x.b, x.c); printf ("Union Y : %d %s %f\n", y.a, y.b, y.c); }
#includemain() { struct Data { int a; int b; } y[4] = { 1, 10, 3, 30, 2, 20, 4, 40}; struct Data *x = y; int i; for(i=0; i<4; i++) { x->a = x->b, ++x++->b; printf("%d %d\t", y[i].a, y[i].b); } }
#includemain() { typedef struct { int a; int b; int c; char ch; int d; }xyz; typedef union { xyz X; char y[100]; }abc; printf("sizeof xyz = %d sizeof abc = %d\n", sizeof(xyz), sizeof(abc)); }
#include#include #define Error(str) printf("Error : %s\n", str); exit(1); main() { int fd; char str[20] = "Hello! Test me"; if ((fd = open("xx", O_CREAT | O_RDWR)) < 0) Error("open failed"); if (write(fd, str, strlen(str)) < 0) Error("Write failed"); if (read(fd, str, strlen(str)) < 0) Error("read failed"); printf("File read : %s\n", str); close(fd); }
#includemain() { int *a, i; a = (int *) malloc(10*sizeof(int)); for (i=0; i<10; i++) *(a + i) = i * i; for (i=0; i<10; i++) printf("%d\t", *a++); free(a); }
Write a program to calculate number of 1's (bit) in a given integer number i.e) Number of 1's in the given integer's equivalent binary representation.
Answers
The output will be : 0 10 11 20 21 *s++ => *(s++) *++s => *(++s) ++*s => ++(*s)
Some compiler (ansi) may give warning message, but it will compile without errors. The output will be : 0 1 2 3 4 and -1
This is the problem about Unions. Unions are similar to
structures but it differs in some ways. Unions can be
assigned only with one field at any time. In this case,
unions x and y can be assigned with any of the one field a
or b or c at one time. During initialisation of unions it
takes the value (whatever assigned ) only for the first
field. So, The statement y = {100} intialises the union y
with field a = 100.
In this example, all fields of union x are assigned with
some values. But at any time only one of the union field
can be assigned. So, for the union x the field c is
assigned as 21.50.
Thus, The output will be
Union 2 : 22 22 21.50
Union Y : 100 22 22
( 22 refers unpredictable results )
The pointer x points to the same location where y is stored. So, The changes in y reflects in x. The output will be : 10 11 30 31 20 21 40 41
The output of this program is purely depends on the
processor architecuture. If the sizeof integer is 4 bytes
and the size of character is 1 byte (In some computers), the
output will be
sizeof xyz = 20 sizeof abc = 100
The output can be generalized to some extent as follows,
sizeof xyz = 4 * sizeof(int) + 1 * sizeof(char) +
padding bytes
sizeof abc = 100 * sizeof(char) + padding bytes
To keep the structures/unions byte aligned, some padding
bytes are added in between the sturcture fields. In this
example 3 bytes are padded between ' char ch' and 'int d'
fields. The unused bytes are called holes. To understand
more about padding bytes (holes) try varing the field types
of the structures and see the output.
Just try to execute this file as such. You can find out
that it will exit immediately. Do you know why?
With this hint, we can trace out the error. If you look
into the macro 'Error', you can easily identify that there
are two separete statements without brases '{ ..}'. That is
the problem. So, it exits after the calling open(). The
macro should be put inside the brases like this.
#define Error(str) { printf("Error : %s\n", str); exit(1); }
This program will fault (Memory fault/segmentation fault). Can you predict Why? Remove the statment 'free(a);' from the program, then execute the program. It will run. It gives the results correctly. What causes 'free(a)' to generate fault? Just trace the address location of pointer variable 'a'. The variable 'a' is incremented inside the 'for loop'. Out side the 'for loop' the variable 'a' will point to 'null'. When the free() call is made, it will free the data area from the base_address (which is passed as the argument of the free call) upto the length of the data allocated previously. In this case, free() tries to free the length of 10 *sizeof(int) from the base pointer location passed as the argument to the free call, which is 'null' in this case. Thus, it generates memory fault.