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We should not read after a write to a file without an intervening call to fflush(), fseek() or rewind() < TRUE/FALSE>
True
Sandhya.Kishan
- Mar 28th, 2012
True, we should not be able to read a file after writing in that file without calling the given functions because if the file was open for writing and IF the last operation was an output operation, th...
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If the program (myprog) is run from the command line as myprog 1 2 3 , What would be the output? main(int argc, char *argv[]) { int i; for(i=0;i<argc;i++) printf("%s",argv[i]); } a) 1 2 3 b) C:MYPROG.EXE 1 2 3 c) MYP d) None of the above
kiran
- May 21st, 2006
The answer is b) C:myprog.exe 1 2 3 . the relative path of the file is passed as the first argument for main method by default.then followed by others.
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On combining the following statements, you will get char*p; p=malloc(100); a) char *p= malloc(100) b) p= (char*)malloc(100) c) All of the above d) None of the above
krishan pratap
- Mar 27th, 2006
The answer would be Because the Their is a need of the decleration the this is the character type of the pointer.
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}
">Vii)7. main(){ int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m);}
0 0 1 3 1Explanation :Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination-...
abhimanipal
- Jan 30th, 2010
Although the answer posted above is correct. There is one mistake in the explanation. Beena says that && operator has higher precedence tha || operator so && is evaluated first. But in...
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case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
">Ix) main(){ int i=3; switch(i) { default:printf("zero"); case 1: printf("one"); break; case 2:printf("two"); break; case 3: printf("three"); break; } }
threeExplanation :The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
Lakshmi Prabha
- Oct 22nd, 2005
The output will be three because when execution takes place ,inside the switch the corresponding case 3(since i=3)is selected and as there is a break statement after it, it cmes out of the loop.
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Diffrence between a "assignment operator" and a "copy constructor"
siva_shiva
- Jun 29th, 2008
Copy constructor intialize the variable and copies from existed object while Assingnment operator uses intialized object.
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What is the purpose of main()function?
Girish.T
- Sep 23rd, 2006
That is from where the compiler starts the execution of the program
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Write a program to print the given integers in the corresponding wordings?
Say for eg. if 123 is given as input the output should be displayed as
sriranga.ch
- Sep 25th, 2008
Main() { int num=123; char c;while(num>1){c=num%10;num=num/10;printf("%c",c);}}
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}
void func2(int a[][10])
{
printf("Will this work?");
}
main()
{
int a[10][10];
func1(a);
func2(a);
}
">Void func1(int (*a)[10]){ printf("Ok it works");}void func2(int a[][10]){ printf("Will this work?");} main() { int a[10][10]; func1(a); func2(a);}
A) Ok it worksWill this work?B) Will this work?C) Ok it worksD) None of the above
visitor
- Mar 3rd, 2006
Ok it works Will this work?
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}
{
unsigned int bit=512;
printf("%d", bit);
}
}
">Main(){ { unsigned int bit=256; printf("%d", bit); } { unsigned int bit=512; printf("%d", bit); }}
A) 256, 512B) 512, 512C) 256, 256D) Compile errorExplanation: because these are different blocks, so declaration is allowed.
supriya ahire
- Mar 21st, 2006
hi, o/p is a)256 512 The program executes block by block.Regards,Supriya Ahire
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What would be the output of the following program? main() { int y=128; const int x=y; printf("%d",x); }
A) 128B) Garbage valueC) ErrorD) 0
Amit
- Dec 12th, 2005
Correct ans is A) 128
C Interview Questions
Ans