STMicroelectronics C Questions: C Questions Part IVTab.amp

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Explanation:

In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e., p = s + 3. The next statement increment value in p by 1, thus now value of p = s + 2. In the printf statement the expression is evaluated * + + p causes gets value s + 1 then the pre decrement is executed and we get s + 1 1 = s. The indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ck.

  1. main () { int i, n; char * x = girl; n = strlen (x); * x = x[n]; for i = 0; i { printf (%s\n, x); x + +; } } Answer: (blank space) irl rl l Explanation: Here a string (a pointer to char) is initialized with a value girl. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (\0') to the first location. Now the string becomes \0irl. Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = \0 hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e.irl and the third time it prints rl and the last time it prints l and the loop terminates.

  2. int i, j; for (i = 0; i ⇐ 10; i + + ) { j + = 5; assert (i<5); } Answer: Runtime error: Abnormal program termination. Assert failed (i<5), Explanation: Asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use, #undef NDEBUG and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of:

  3. main () { int i = -1; + i; printf ( “i = %d, + i = %d \n” i, + i); } Answer: i = -1, + i = -1 Explanation: Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

  4. What are the files which are automatically opened when a C file is executed? Answer: Stdin, stdout, stderr (standard input, standard output, standard error).

  5. what will be the position of the file marker? a: Fseek (ptr, 0, SEEK_SET); b: Fseek (ptr, 0, SEEK_CUR); Answer: a: The SEEK_SET sets the file position marker to the starting of the file. b: The SEEK_CUR sets the file position marker to the current position of the file.

  6. main () { char name[10], s[12]; scanf ( “\” %[^\ “]\”", s); } How scanf will execute? Answer: First it checks for the leading white space and discards it. Then it matches with a quotation mark and then it reads all character upto another quotation mark.

  7. What is the problem with the following code segment? while ( (fgets (receiving array, 50, file_ptr) )! = EOF); Answer & Explanation: Fgets returns a pointer. So the correct end of file check is checking for! = NULL.

  8. main () { main (); } Answer: Runtime error: Stack overflow. Explanation: Main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

  9. main () { char * cptr, c; void * vptr, v; c = 10; v = 0; cptr = &c; vptr = &v; printf ( “%c%v” c, v); } Answer: Compiler error (at line number 4): Size of v is Unknown. Explanation: You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

  10. main () { char * str1 = “abcd” char str2[] = “abcd” printf ( “%d %d %d” sizeof (str1), sizeof (str2), sizeof ( “abcd” ) ); } Answer: 2 5 5 Explanation: In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the ‘\0’ termination character). The third sizeof is similar to the second one.

  11. main () { char not; not =! 2; printf ( “%d” not); } Answer: 0 Explanation: is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. TRUE is FALSE (0) so it prints 0.

  12. #define FALSE-1 #define TRUE 1 #define NULL 0 main () { if (NULL) puts ( “NULL” ); else if (FALSE) puts ( “TRUE” ); else puts ( “FALSE” ); } Answer: TRUE Explanation: The input program to the compiler after processing by the preprocessor is, main () { if (0) puts ( “NULL” ); else if (-1) puts ( “TRUE” ); else puts ( “FALSE” ); } Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if-1 is boolean value true hence “TRUE” is printed.

  13. main () { int k = 1; printf ( “%d = = 1 is” “%s” k, k = = 1? “TRUE” “FALSE” ); } Answer: 1 = = 1 is TRUE Explanation: When two strings are placed together (or separated by white-space) they are concatenated (this is called as “stringization” operation). So the string is as if it is given as “%d = = 1 is %s” The conditional operator (? ) evaluates to “TRUE”

  14. main () { int y; scanf ( “%d” &y);//input given is 2000 if ( (y%4 = = 0 && y%100! = 0) | y%100 = = 0) printf ( “%d is a leap year” ); else printf ( “%d is not a leap year” ); } Answer: 2000 is a leap year Explanation: An ordinary program to check if leap year or not.

  15. #define max 5 #define int arr1[max] main () { typedef char arr2[max]; arr1 list = { 0, 1, 2, 3, 4}; arr2 name = “name” printf ( “%d %s” list[0], name); } Answer: Compiler error (in the line arr1 list = { 0, 1, 2, 3, 4}) Explanation: Arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error. Rule of Thumb: #defines are used for textual replacement whereas typedefs are used for declaring new types.

  16. int i = 10; main () { extern int i; { int i = 20; { const volatile unsigned i = 30; printf ( “%d” i); } printf ( “%d” i); } printf ( “%d” i); } Answer: 30, 20, 10 Explanation: Introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

  17. #include main () { FILE * ptr; char i; ptr = fopen ( “zzz. c” “r” ); while ( (i = fgetch (ptr) )! = EOF) printf ( “%c” i); } Answer: Contents of zzz. c followed by an infinite loop

  18. main () { int i = -1; i; printf ( “i = %d, i = %d \n” i, i); } Answer: i = -1, i = 1 Explanation: It is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression-i = - (-1) is printed.

  19. #include main () { const int i = 4; float j; j = + + i; printf ( “%d %f” i, + + j); } Answer: Compiler error Explanation: i is a constant. You cannot change the value of constant

  20. #include main () { int a[2][2][2] = { { 10, 2, 3, 4}, { 5, 6, 7, 8} }; int * p, * q; p = &a[2][2][2]; * q = * * * a; printf ( “%d. %d” * p, * q); } Answer: Garbagevalue. 1 Explanation: p = &a[2][2][2] you declare only two 2D arrays. But you are trying to access the third 2D (which you are not declared) it will print garbage values. * q = * * * a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print * q meAnswer: It will print first element of 3D array.

  21. #include main () { register i = 5; char j[] = “hello” printf ( “%s %d” j, i); } Answer: Hello 5 Explanation: If you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

  22. main () { int i = 5, j = 6, z; printf ( “%d” i + + + j); } Answer: 11 Explanation: The expression i + + + j is treated as (i + + + j)

  23. struct point { int x; int y; }; struct point origin, * pp

  24. main () { int i = _l_abc (10); printf ( “%d\n” -i); } int _l_abc (int i) { return (i + + ); } Answer: 9 Explanation: Return (i + + ) it will first return i and then increments. Ie, 10 will be returned.

  25. main () { char * p; int * q; long * r; p = q = r = 0; p + +; q + +; r + +; printf ( “%p…%p…%p” p, q, r); } Answer: 0001…0002…0004 Explanation: + + operator when applied to pointers increments address according to their corresponding data-types.

  26. main () { char c = ‘’ x, convert (z); getc (c); if ( (c> = ‘a’ ) && (c ⇐ ‘z’ ) ) x = convert (c); printf ( “%c” x); } convert (z) { return z − 32; } Answer: Compiler error Explanation: Declaration of convert and format of getc () are wrong.

  27. main (int argc, char * * argv) { printf ( “enter the character” ); getchar (); sum (argv[1], argv[2]); } sum (num1, num2) int num1, num2; { return num1 + num2; } Answer: Compiler error. Explanation: Argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

  28. # include int one_d[] = { 1, 2, 3}; main () { int * ptr; ptr = one_d; ptr + = 3; printf ( “%d” * ptr); } Answer: Garbage value Explanation: Ptr pointer is pointing to out of the array range of one_d.

  29. # include aaa () { printf ( “hi” ); } bbb () { printf ( “hello” ); } ccc () { printf ( “bye” ); } main () { int ( * ptr[3]) (); ptr[0] = aaa; ptr[1] = bbb; ptr[2] = ccc; ptr[2] (); } Answer: Bye Explanation: Ptr is array of pointers to functions of return type int. Ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. Ptr[2] () is in effect of writing ccc (), since ptr[2] points to ccc.

  30. main () { int * j; { int i = 10; j = &i; } printf ( “%d” * j); }

Answer: 10

Explanation: The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, * j prints the value stored in i since j points i.