48 lines
1.3 KiB
Markdown
48 lines
1.3 KiB
Markdown
> 2. The following program fragments illustrate the logical operators. Show the output produced by each, assuming that `i`, `j`, and `k` are `int` variables.
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a. `i = 10; j = 5;`
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```c
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printf("%d", !i < j);
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// Expected output: `1`, because `!i` evaluates to 0, and 0 is less than 5, so that expression evaluates to true, or 1.
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```
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b. `i = 2; j = 1;`
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```c
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printf("%d", !!i + !j);
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// Expected output: `1`, because !!2 evaluates to 1, and !j evaluates to 0
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```
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c. `i = 5; j = 0; k = -5;`
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```c
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printf("%d", i && j || k);
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// Expected output: `1`, because i && j should evaluate to 0, but `0 || 1` should evalulate to true.
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```
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d. `i = 1; j = 2; k = 3;`
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```c
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printf("%d", i < j || k);
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// Expected output: `1`
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```
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> 4. Write a single expression whose value is either `-1`, `0`, or `1` depending on whether `i` is less than, equal to, or greater than `j`, respectively.
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```c
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/*
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If i < j, the output should be -1.
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If i == j, the output should be zero
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If i > j, the output should be 1.
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*/
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(i > j) - (i < j)
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```
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> 6. Is the following `if` statement legal?
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```c
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if (n >= 1 <= 10)
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printf("n is between 1 and 10\n");
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```
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Yes the statement is *legal*, but it does not produce the intended effect. It would produce an output when `n` is zero because `0 >= 1` would evaluate to `0`, and `0 <= 10` is false. |