Tuesday, February 16, 2010
Saturday, February 13, 2010
c,c++ qns
22. What will the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
c,c++ qns
19.
void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"
20. Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
else
printf("Fail2");
}
else
printf("Fail1);
if(i
else
printf("Fail2")
}
a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"
20. Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
else
printf("Fail2");
}
else
printf("Fail1);
if(i
else
printf("Fail2")
}
a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
c,c++ qns
15.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}
Ans: 20 20 20
16. There was question in c working only on unix machine with pattern matching.
14. what is alloca()
Ans : It allocates and frees memory after use/after getting out of scope
17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}
Ans: 321
18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}
Ans: anything is good.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}
Ans: 20 20 20
16. There was question in c working only on unix machine with pattern matching.
14. what is alloca()
Ans : It allocates and frees memory after use/after getting out of scope
17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}
Ans: 321
18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}
Ans: anything is good.
c,c++ qns
10. A structure pointer is defined of the type time . With 3 fields min,sec hours having pointers to intergers.
Write the way to initialize the 2nd element to 10.
11. In the above question an array of pointers is declared.
Write the statement to initialize the 3rd element of the 2 element to 10;
12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}
What are the number of syntax errors in the above?
Ans: None.
13.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");
Ans: "one is defined"
Write the way to initialize the 2nd element to 10.
11. In the above question an array of pointers is declared.
Write the statement to initialize the 3rd element of the 2 element to 10;
12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}
What are the number of syntax errors in the above?
Ans: None.
13.
void main()
{
int i=7;
printf("%d",i++*i++);
}
Ans: 56
14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");
Ans: "one is defined"
c,c++ qns
8.
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}
1. 1000000
2. Overflow
3. Error
4. None
Ans: 4
9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
Ans: The first part of this question can be answered in at least
three ways:
1. char *(*(*a[N])())();
2. Build the declaration up incrementally, using typedefs:
typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */
3. Use the cdecl program, which turns English into C and vice
versa:
cdecl> declare a as array of pointer to function returning
pointer to function returning pointer to char
char *(*(*a[])())()
cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these complicated
C declarations "inside out" to understand them ("declaration
mimics use").
The pointer-to-function declarations in the examples above have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}
1. 1000000
2. Overflow
3. Error
4. None
Ans: 4
9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
Ans: The first part of this question can be answered in at least
three ways:
1. char *(*(*a[N])())();
2. Build the declaration up incrementally, using typedefs:
typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */
3. Use the cdecl program, which turns English into C and vice
versa:
cdecl> declare a as array of pointer to function returning
pointer to function returning pointer to char
char *(*(*a[])())()
cdecl can also explain complicated declarations, help with
casts, and indicate which set of parentheses the arguments
go in (for complicated function definitions, like the one
above).
Any good book on C should explain how to read these complicated
C declarations "inside out" to understand them ("declaration
mimics use").
The pointer-to-function declarations in the examples above have
not included parameter type information. When the parameters
have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)
C,C++ Questions
5.
void main()
{
int i;
for(i=1;i<4,i++)
switch(i)
case 1: printf("%d",i);break;
{
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}
Ans: 1,2,3,4
6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}
Ans: 6
7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}
Ans: less
void main()
{
int i;
for(i=1;i<4,i++)
switch(i)
case 1: printf("%d",i);break;
{
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}
Ans: 1,2,3,4
6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}
Ans: 6
7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(i
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}
Ans: less
C,C++ Questions
C,C++ Questions
1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived
object,. calling of that virtual method will result in which method being called?
a. Base method
b. Derived method..
Ans. b
2. For the following C program
#define AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}
What is the output?
Ans. Area of the circle is 122.656250
Area of the circle is 19.625000
3. What do the following statements indicate. Explain.
· int(*p)[10]
· int*f()
· int(*pf)()
· int*p[10]
Refer to:
-- Kernighan & Ritchie page no. 122
-- Schaum series page no. 323
4.
void main()
{
int d=5;
printf("%f",d);
}
Ans: Undefined
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mobile printing on hp
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Quantum computing
the quantum computing is the fastest growing technology
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mobile computing
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Friday, February 12, 2010
BIOMEDICAL INSTRUMENTATION-1
AUTOMATED TEST SYSTEM FOR FIBER OPTIC ENDOSCOPES USING
VIRTUAL INSTRUMENTATION
VIRTUAL INSTRUMENTATION BASED AUTOMATED TEST SYSTEM FOR FIBER OPTIC ENDOSCOPES
ABSTRACT:
This paper deals with the application of Virtual Instrumentation in the field of Bio-Medical Instrumentation. The VI based Endo-Tester is used to assess the optical performance of the fiber optic endoscopes and video system in a clinical setting. Using LabVIEW and IMAQ products we can build an integrated test bench suitable for use by the clinical staff to measure parameters such as light loss, symmetry and distortion in the Fiber Optic Endoscopes.
INTRODUCTION
The popularity of endoscopic surgery is growing, in large part because it is generally safer and less expensive than conventional surgery. As endoscopic surgery becomes more common, accurately evaluating the performance of endoscopes and their peripheral components becomes essential. Assessing the optical performance of endoscopes and video systems is often difficult in the clinical setting. The surgeon depends on a high-quality image to perform minimally invasive surgery, yet assurance of proper function of the equipment by biomedical engineering staff is not always straight forward. Poor images can result from many variables, such as optical problems with the rigid endoscope, damaged fibers in the light cable, video camera malfunctions, and cleanliness of the equpment, lens surfaces on both ends of the endoscope.
We proposed an idea to develop the EndoTester evaluation system in response to worldwide demand for an affordable, accurate and highly versatile test system for assessing the performance characteristics of endoscopes and video systems in the clinical setting.
THE ENDO TESTER EVALUATION SYSTEM
The Endo Tester is an easy-to-use system to assess and accurately quantify the optical characteristics of fiber optic endoscopes. The system which we are going to design from its conception for ease of use, accommodates virtually any type of fiber optic endoscope. Multimedia features and on-line help and instructions provide support for biomedical engineering staff, nurses and physicians. By leveraging off standard PCs and an IMAQ image acquisition board, this LabVIEW based system is highly versatile and affordable.
VIRTUAL INSTRUMENTATION
VIRTUAL INSTRUMENTATION BASED AUTOMATED TEST SYSTEM FOR FIBER OPTIC ENDOSCOPES
ABSTRACT:
This paper deals with the application of Virtual Instrumentation in the field of Bio-Medical Instrumentation. The VI based Endo-Tester is used to assess the optical performance of the fiber optic endoscopes and video system in a clinical setting. Using LabVIEW and IMAQ products we can build an integrated test bench suitable for use by the clinical staff to measure parameters such as light loss, symmetry and distortion in the Fiber Optic Endoscopes.
INTRODUCTION
The popularity of endoscopic surgery is growing, in large part because it is generally safer and less expensive than conventional surgery. As endoscopic surgery becomes more common, accurately evaluating the performance of endoscopes and their peripheral components becomes essential. Assessing the optical performance of endoscopes and video systems is often difficult in the clinical setting. The surgeon depends on a high-quality image to perform minimally invasive surgery, yet assurance of proper function of the equipment by biomedical engineering staff is not always straight forward. Poor images can result from many variables, such as optical problems with the rigid endoscope, damaged fibers in the light cable, video camera malfunctions, and cleanliness of the equpment, lens surfaces on both ends of the endoscope.
We proposed an idea to develop the EndoTester evaluation system in response to worldwide demand for an affordable, accurate and highly versatile test system for assessing the performance characteristics of endoscopes and video systems in the clinical setting.
THE ENDO TESTER EVALUATION SYSTEM
The Endo Tester is an easy-to-use system to assess and accurately quantify the optical characteristics of fiber optic endoscopes. The system which we are going to design from its conception for ease of use, accommodates virtually any type of fiber optic endoscope. Multimedia features and on-line help and instructions provide support for biomedical engineering staff, nurses and physicians. By leveraging off standard PCs and an IMAQ image acquisition board, this LabVIEW based system is highly versatile and affordable.
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