scanf(或) printf 中的格式说明符错误 [英] Wrong format specifiers in scanf (or) printf

问题描述

就printf函数而言，我从一些参考资料和实验中了解到以下内容.

As of the printf function is concerned, I understand the following from few references and experiments.

• 当我们尝试使用用于浮点(或)双精度的格式说明符打印整数值时，反之亦然，行为是不可预测的.
• 但是可以使用 %c 打印与整数值等效的字符.也可以使用 %d 打印字符的 ASCII 值(整数表示).

• When we try to print an integer value with format specifiers that are used for float (or) double and vice the versa the behaviour is unpredictable.
• But it is possible to use %c to print the character equivalent of the integer value. Also using of %d to print ASCII value (integer representations) of character is acceptable.

同样，如果格式说明符和传递给 scanf 的参数不匹配，scanf 的行为是什么.标准是否定义了它?

Similarly, what is the behaviour of scanf, if there is a mismatch of format specifier and the arguements passed to scanf. Does the standards define it?

推荐答案

可变参数(那些匹配省略号，...)是默认提升的.这意味着所有较短的整数类型都被提升为 int(或无符号，视情况而定).整数和字符之间没有区别(我相信).printf 中 %d 和 %c 之间的区别仅在于值的格式.

Variadic arguments (those matching the ellipsis, ...) are default-promoted. That means that all shorter integral types are promoted to int (or unsigned, as appropriate). There's no difference between inte­gers and characters (I believe). The difference between %d and %c in printf is merely how the value is formatted.

scanf 是另一种鱼.您传递的所有参数都是指针.指针之间没有默认的运动，并且传递与指针类型匹配的确切格式说明符至关重要.

scanf is a different kettle of fish. All the arguments you pass are pointers. There's no default-pro­mo­tion among pointers, and it is crucial that you pass the exact format specifier that matches the type of the pointee.

在任何一种情况下，如果您的格式说明符与提供的参数不匹配(例如将 int * 传递给 printf 中的 %p>)，结果是未定义行为，这远比不可预测"更糟糕——这意味着你的程序只是格式错误.

In either case, if your format specifier doesn't match the supplied argument (e.g. passing an int * to a %p in printf), the result is undefined behaviour, which is far worse than being "unpredictable" -- it means your program is simply ill-formed.

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