【程序设计与算法基础】选/判/填 抽象Trick集

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if(x > 0) y = 1;
else if(x < 0) y = -1;
else if(x = 0) y = 0;

Will the program run as expected?

Answer is NO.

Here is a list of stupid tricks in C, which may make you stumble.

0<=x<=10

显然C不支持这种连等式。这段代码可以理解成((0<=x)?1:0)<=10.

对于==,>=,<,>同理

if(x = 0)

[Warning] suggest parentheses around assignment used as truth value [-Wparentheses]

小心if中出现的这种抽象副作用。

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int x,y=0;
if( x = 0 ){
    y = 1;
} else {
    y = 2;
}

x,y的值是多少？

由于x未定义，值未知，所以y也未知？并非如此，x被赋值为0,y的值为2.

if走那条分支取决于赋值后的x,如果x是Falsy的(e.g. 0,false)，if会走false支.

Where is your semicolon ;

第一空填什么

填t = w / (h * h)？ 废了，分号没辣！

Where is your parentheses ()

同上，程序填空题喜欢动不动就把if和for左右的括号漏掉。

以下是翻-Wall文档时找到的一些东西，这些暂时没遇到过

数组下标用char类型

-Wchar-subscripts

隐式精度提升

-Wdouble-promotion

Give a warning when a value of type float is implicitly promoted to double. CPUs with a 32-bit “single-precision” floating-point unit implement float in hardware, but emulate double in software. On such a machine, doing computations using double values is much more expensive because of the overhead required for software emulation.
当 float 类型的值隐式提升为 double 时发出警告。具有 32 位“单精度”浮点单元的 CPU 在硬件中实现 float ，但在软件中模拟 double 。在这样的机器上，使用 double 值进行计算要昂贵得多，因为软件仿真所需的开销。

It is easy to accidentally do computations with double because floating-point literals are implicitly of type double. For example, in:
很容易意外地使用 double 进行计算，因为浮点字面量隐式为 double 类型。例如，在：

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float area(float radius)
{
   return 3.14159 * radius * radius;
}

the compiler performs the entire computation with double because the floating-point literal is a double.
编译器使用 double 执行整个计算，因为浮点字面量是 double。

switch fallthrough

-Wimplicit-fallthrough

The switch will fallthrough.

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switch (cond)
  {
  case 1:
    a = 1;
    break;
  case 2:
    a = 2;
  case 3:
    a = 3;
    break;
  }

So do the code below:

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switch (cond)
  {
  case 1:
    if (i > 3) {
      bar (5);
      break;
    } else if (i < 1) {
      bar (0);
    } else
      return;
  default:
    …
  }

What is between { and the first case

What is between the controlling expression and the first case label?

-Wswitch-unreachable

When a switch statement contains statements between the controlling expression and the first case label, it will never be executed.

Except when it is a declaration.

And in deed…

What is the lifecycle of vars in switch

Varriables can be declared but not be initialized.

Once declared, it can be used across the scope.

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switch (cond) {
		int u;
	case 5:
		i = 5;
		int c;
		break;
	case 7:
		c=1;
		cout<<c;
	case 8:
		u=9;
		break;
}

It can be compiled and run as expected, even it seems c is not declared when cond==7 and u seems to be never declared.

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switch (cond) {
		int u=1;
	case 5:
		i = 5;
		int c=4;
		break;
	case 7:
		c=1;
		cout<<c;
	case 8:
		c=9;
		break;
}

And when you try to initialize the variables, compilers start to stop you with [Error] crosses initialization of 'int c'

To be continued.