C2

THE TEST ON C

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(1)

Explain the output of the following program. #define FALSE -1 #define TRUE 1 #define NULL 0 int main() { if(NULL) puts("NULL"); else if(FALSE) puts("TRUE"); else puts("FALSE"); } ?

output is : TRUE EXPLANATION: at step : if(NULL) evaluates to --if(0);so if is not executed then at next step : else if(FALSE) evaluates to -- if(-1);so will execute it and execute the line after it: puts("TRUE"); --o/p = TRUE then at next step : else --will not execute after executing else if statement

PR Attribute ID Marks New MarksC_1 1 1C_5 2 2

(2) Place holder for Large Question (To de done on server) ?

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2/6/2009http://10.203.161.13/OES/take1.jsp?s1=504&s2=20429

PR Attribute ID Marks New MarksC_12 2 1.5C_14 2 2C_17 4 3C_19 6 3.5C_20 2 1.5

(3)

void f2(int); void f1(int a) { if( a ) f2( a - 1 ); printf("%d", a); } void f2(int b) { printf("."); if(b) f1( b - 1 ); } int main(void) { f1(5); }

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Explain the output of this program. ?

the output is => ...135 i.e 3 dots followed by 1 3 and 5 EXPLANATION => IN MIAN call to f1(5) now at f1: void f1(5) { if(5) f2(4); ------------->void f2(4) printf("%d",a); ''here a = 5 { printf("."); //o/p's a '.' } if(4) f1(3); ---> void f1(3) } { if(3) f2(2) -------------------------- printf("%d",a); //here a = 3 } void f2(2) <-------------------- { printf(".");//o/p's a '.' if(2) void f1(1) <------ f1(1); { } if(1) f2(0) --->void f2(0) //a = 1 printf("%d",a) { printf(".")//o/p '.' if(0) f(0) }

PR Attribute ID Marks New MarksC_8 4 4

(4)

int i =0; void f1(void) { static int k; i = 50; printf("%d\n",k); } int main() { int j; f1(); i =0; printf(" i in main %d\n",i); f1(); printf(" i after call%d\n",i); } Explain the output of this program? ?

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OUTPUT IS: 0 i in main 0 0 i after call 0 EXPLANATION IS : IN MIAN { first call to f1()--> will initialise a static variable k = 0 then i = 50 then prints k as--> o then in main-->prints --> i in main 0 (as i is defined in its local scope also therefore value of is 0 then calls to f1() again--> will prints k as no change in it -- then back in mian--> will print--> i after call 0(as no change in i local to it


(5)

What would be output of this program? What are the potential issues with respect to the followng piece of code?int main() { FILE * fp = fopen("test.txt", "r"); char line[100]; while( ! feof(fp) ) { fgets(line, sizeof(line), fp); fputs(line, stdout); } } Note: Assume that a file "test.txt" exists and the content is abc def ?

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OUTPUT IS --> abc def def EXPLANATION --> AT FIRST fgets reads line:abc and prints to console then in loop again get line def from file and prints to console then feof will return zero this time also and control moves to loop again and prints def again,after that loop ends as feof() returns non to resolve this issue : while(1) { fgets(line,sizeof(line),fp); if(feof(fp)) break; fputs(line,stdout); } this will print like: abc def


(6)

#define MAX 10 void fun1(int data[]) { int index; for (index = 0; index < MAX; ++index) data[index] = 0; } void fun2(int *data_ptr) { int index; for (index = 0; index < MAX; ++index) *(data_ptr + index) = 0; } What are fun1() and fun2() trying do? What is the difference between them? ?

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BOTH FUNCTIONS ARE INITIALISING THE ARRAY,ONE OF THEM IS ACCESSING IT AS AN ARRAY OF INTEGERS AND OTHER IS ACCESSING IT AS POINTER TO INTEGER (WILL BEHAVE AS ARRAY OF INTEGERS) THE FUNCTION FUN1 IS TAKING AN INTEGER ARRAY AS DATA[] AND INITALISING IT AS DATA[i] BY FOR LOOPAND THE FUNCTION fun2 IS TAKING A INTEGRE POINTER AS *DATA_PTR AND INITIALISING IT BY MOVING THROUGHTHE INDEXES


(7)

For which input the following program would print YES? Explain your answer. int main() { int a; printf( "Enter the number to be tested: " ); scanf( "%d", &a ); if(a & ( a >> 1 ) == 0 ) { printf( "YES"); } else { printf( "NO"); } } ?

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IT WOULD PRINT YES FOR THE INPUTS LIKE--> 1,2,4,5,8,9,10 AS HERE IF STATEMENT IS CHECKING FOR ZERO IN EXPRESSION-->(a & (a>>1)) i.e a IS AND WITH (a AFTER SHIFTING IT BY ONE BIT) THUS ALL THE NUMEBRS IN BINARY WHICH AFTER SHIFTING BY ONE BIT WILL NOT HAVE THE ONE BIT IN THE SAME POSITIONS OF EARLIE 1'S WILL EVALUATE TO ZERO AND THEN IF STATEMENT WILL GIVE RESULT AS 1 AND PRINTS


(8)

char takes 1 byte, int takes 4 bytes, double takes 8 bytes. struct S1 { char c; int i[2]; double v; } SA1; struct S2 { double x; int i[2]; char c; } SA2; What will be the size of these structures? ?

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IF THE MACHINE IS 64 BIT MACHINE: THEN A WORD IS 8 BYTE LONG IN 64 BIT MACHINE SIZE OF --> SA1 IS: 24 BYTES -->SA2 IS 24 BYTES EXPLANATION IS: IN 8 BYTE WORD FOR SA1 AT FIRST CHAR c is stored and the for stroing int i[0] will start from from byte:4 (a multiple of 4) thus occuping one word the i[1] is stroed in next 4 bytes of next word thus occupying the 2nd word then double v will start from next word as needs to be stroed in 8 multiple address space FOR SA2: AT FIRST WORD DOUBLE OCCYPIES THE 8 BYTES,THE INT IP[0] AND INT I[1] OCYPIES THE NEXT COMPLETE WORD AND THEN CHAR C WILL OCUPPY 1 BYTE ,BUT COMPLIER WILL READ THE COMPLET WORDS AS REST BYTES ARE ALL PADDED in 32 BIT MACHINE : HAS 4 BYTE LONG WORD --> SA1 IS :20 BYTES LONG --> SA2 IS : 20 BYTES LONG EXPLANATION IS : IN 4 BYTE WORD FOR SA1 CHAR C IS STROED IN FIRST WORD,FOR INT i[0] IS STROED IN NEXT WORD AS TO BE STROED IN MULTIPLE OF 4 ,AND REST BYTES FOR FIRST WORD ARE PADDED,THEN I[1] IS STORED IN 3RD WORD AND THEN DOUBLE VOCCUPIS THE NEXT BYTES OF NEXT TWO WORDS THUS OCCUPING THE 20 BYTES FOR SA2: AT FIRST DOUBLE X OCUPPIES THE FIRST 2 WORDS,THEN INT I[0] AND I[1] OCUPPIES THE NEXT TWO WORDS AND THE CHAR C OCCUPY FIRST BYTE OF 5TH WORD BUT REST ALL BYTES ARE PADDED,THUS OCCUPYING 20 BYTES AS A WHOLE


(9)

int main() { int i; for (i = 0; i < 3; ++i) { int t = 1; static int i = 1; int *ptr = &i; printf("%d %d\n",t, *ptr); t = t + 1; *ptr = *ptr + 1; } } Explain the output of this program. ?

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OUTPUT IS : 11 12 13 EXPLANATION IS=> AS IN FOR LOOP { int t = 0 is defined with auto stroge type and made equals to 0 each time of the for loop thus its value will persist in different for lopp execution as variable i is define as static thus i's value will persist between different for loop execution. as *ptr is pointing to i theerfore i is increades by one by statemnet *ptr = *ptr + 1 for(i = 0;i < 3;++i) //i =0 { -->prints : t as 1 and *ptr as 1 the again for( ) //--> i =1 -->prints : t as 0 and *ptr as 2 as t will be initalised again an again and i same as earlier call then in next for( -->i = 2 -->prints : t as 0 and *ptr is 2 then in agian for loop i becomes 3 therfore exits


(10)

int main() { int i,j; int ctr = 0; int myArray[2][3]; for (i=0; i< 2; i++) for (j=0; j< 3; j++) { myArray[i][j] = ctr; ++ctr; } for(i=0;i<3;i++) for(j=0;j<2;j++) printf("%d\t",*(*(myArray+i)+j)); } Explain the behaviour of this program. What is the bug here? ?

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in main: A ARRAY IS INITALISED WITH 2 ROWS AND THREE COLUMNS AT FIRST FOR LOOPS WE ACCESSING ARRAY ELEMENTS AND INITIALISING THEM WITH CTR VALUES INCREADES EACH TIME AND IN SECOND FOR LOOPS WE ARE DISPLAYING THEM BUT ACCESSING THEM BY AT FIRST THREE ROWS AND THEN IN INNER LOOP TWO COLUMNS AR TRAVERSED THIS PROGRAM WILLL BHEAVE NORMALLY AS MEMORY FOR ARRAY ELEMENT IS STORED IN CONTIGUOUS LOCATIONS THEREORE EACH ELEMENT IS ACCESED IN THE SECOND FOR LOOPS ALSO AS MYARRAY[0][0] -->MYARRAY[0][0] MYARRAY[0][1] --. MYARRAY[0][1] MYARRAY[0][2]--> MYARRAY[1][0] MYARRAY[1][0] -->MYMARRAY[1][1] MYARRAY[1][1] --> MYARRAY[2][0] MYARRAY[1][2]-->MYARRAY[2][1] ARE ACCESED BY THE SECOND FOR LOOPS THEREFOIRE IS ONLY THAT ARRAY ELEMENTS ARE ACCESSED IN PROPER FASHION AS SHOULD BE


(11)

#define SIZE 20 void function1(void) { char *s1 = "my"; char * s2 = "program"; char *s3 = strcat(s1,s2); printf("%s",s3); } void function2(void) { char *s1 = (char *) malloc(SIZE * sizeof(char)); strcpy(s1,"my"); char *s2 = "program"; char *s3 = strcat(s1,s2); printf("%s\n",s3); } Explain what would happen at function1 and function2? Point out the bugs, if any. ?

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AT FUNCTION1: *s1 ans *s2 and *s3 should be first given memory and then should be assigned to a memory address location not the string,as only strcopy works for copying strings from one to another,and only address or onter pointer can be assigned to another pointerthus steps wise: char *s1 = "my"; will give error char *s2 = "program" //error char *s3 = strcat(s1,s2); will not print s3 at function2 : at first menmory is given to s1 then through strcpy "my" is copied to s1 the char *s2 will give error cahr *s3 = strcat(s1,s2) //error as no s2 will not print s3 bugs here is : at first memory should be assigned to pointers and then strcoy is to be used to copy strings or asigned to some other pinter


(12)

int main(void) { int **pptr,*ptr, q; ptr = &q; pptr = &ptr; q = 1; printf("%p ", ptr); *ptr++; printf("%d %p\n", q, ptr); **pptr++; printf("%d %p\n", q, *pptr); } What would be the output of the above program? Justify your answer. ?

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suppose the address of q is :0x200 therefore output is : 0x200 2 0x200 3 0x200 there will be no spaces in between justification : q = 1 ptr = &q i.e 0x200 (suppose) pptr = &ptr (i.e. *pptr = ox200) therefore first printf prints--> ptr value 0x200 then *ptr++ i.e q= q+1 = 2 then next printf prints --. q as 2 and ptr as 0x200 **pptr++ i.e. q = q + 1 i.e q= 3 then next printf prints --> q as 3 and *pptr as 0x200


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