C中的后增量和前增量

Question

我对这两个C语句有疑问:

1. x = y++;

2. t = *ptr++;

使用语句1,y的初始值被复制到x然后y递增.

使用语句2,我们查看*ptr指向的值,将其放入变量t,然后稍后增加ptr.

对于语句1,后缀增量运算符的优先级高于赋值运算符.所以不应该首先递增y,然后将x赋值给y的递增值？

在这些情况下,我不理解运算符优先级.

Answer 1

C中前自增和后自增的区别：

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预自增和后自增是内置的一元运算符。一元表示：“具有一个输入的函数”。“运算符”的意思是：“对变量进行修改”。

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自增 (++) 和自减 (--) 内置一元运算符修改它们所附加的变量。如果您尝试对常量或文字使用这些一元运算符，您将收到错误。

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在 C 中，以下是所有内置一元运算符的列表：

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Increment:         ++x, x++\nDecrement:         \xe2\x88\x92\xe2\x88\x92x, x\xe2\x88\x92\xe2\x88\x92\nAddress:           &x\nIndirection:       *x\nPositive:          +x\nNegative:          \xe2\x88\x92x\nOnes_complement:  ~x\nLogical_negation:  !x\nSizeof:            sizeof x, sizeof(type-name)\nCast:              (type-name) cast-expression\n

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这些内置运算符是伪装的函数，它们接受变量输入并将计算结果放回到同一变量中。

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后置增量示例：

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int x = 0;     //variable x receives the value 0.\nint y = 5;     //variable y receives the value 5\n\nx = y++;       //variable x receives the value of y which is 5, then y \n               //is incremented to 6.\n\n//Now x has the value 5 and y has the value 6.\n//the ++ to the right of the variable means do the increment after the statement\n

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预增量示例：

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int x = 0;     //variable x receives the value 0.\nint y = 5;     //variable y receives the value 5\n\nx = ++y;       //variable y is incremented to 6, then variable x receives \n               //the value of y which is 6.\n\n//Now x has the value 6 and y has the value 6.\n//the ++ to the left of the variable means do the increment before the statement\n

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后减量示例：

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int x = 0;     //variable x receives the value 0.\nint y = 5;     //variable y receives the value 5\n\nx = y--;       //variable x receives the value of y which is 5, then y \n               //is decremented to 4.\n\n//Now x has the value 5 and y has the value 4.\n//the -- to the right of the variable means do the decrement after the statement\n

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预减示例：

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int x = 0;     //variable x receives the value 0.\nint y = 5;     //variable y receives the value 5\n\nx = --y;       //variable y is decremented to 4, then variable x receives \n               //the value of y which is 4.\n\n//x has the value 4 and y has the value 4.\n//the -- to the right of the variable means do the decrement before the statement\n

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Answer 2

你错了你的意思2].后递增总是从递增之前产生值,然后在某个时间之后递增该值.

因此,t = *ptr++基本上相当于:

t = *ptr;
ptr = ptr + 1;

这同样适用于您1]- 产生y++的值是y增量前的值.优先级不会改变 - 无论表达式中其他运算符的优先级有多高或多少,它产生的值将始终是增量之前的值,并且增量将在之后的某个时间完成.