What is the correct way to round offx, a float, to an int value?
| A. | y=(int)(x+0.5) |
| B. | y=int(x+0.5) |
| C. | y=(int)x+0.5 |
| D. | y=(int)(int)x+0.5) |
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Option: A Explanation : Click on Discuss to view users comments. |
By default, any real number in 'C' is treated as
| A. | A float |
| B. | A double |
| C. | A long double |
| D. | Depend upon memory model that you are using |
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Option: B Explanation : Click on Discuss to view users comments. |
To print out a and b given below, which printf() statement would you use?
float a = 3.14;
double b = 3.14;
| A. | printf("%f%f",a,b); |
| B. | printf("%Lf%f",a,b); |
| C. | printf("%Lf%Lf",a,b); |
| D. | printf("%f%Lf",a,b); |
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Option: A Explanation : Click on Discuss to view users comments. |
When a variable of data type double is converted into float, then
| A. | Rounding takes place |
| B. | Truncation takes place |
| C. | The lower order bits are dropped |
| D. | None of these |
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Option: A Explanation : Click on Discuss to view users comments. |
The library function sqrt operates on a double precision argument. If, i is an integer variable, then which of the following calls would correctly compute sqrt(i)?
| A. | sqrt((double)i) |
| B. | (double)sqrt(i)? |
| C. | (double)(sqrt(i)) |
| D. | sqrt(i) |
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Option: A Explanation : Click on Discuss to view users comments. |
