char指针或char变量的默认值是什么 [英] what is the default value of char pointer or char variable

问题描述

下面是我尝试打印默认值/char变量&的值的代码.指针.但是在控制台上看不到它是否具有默认值或只是无法读取ASCII范围?

#include <stdio.h>
int main()
{
    char c, *cp;

    printf("\nValue of char c:%c\n", c);
    printf("\nValue of char ptr:%c\n", *cp);

    return 0;
}

解决方案

惊喜！这里没有默认"值，您正在尝试未定义的行为.

详细说明，如果变量是局部范围的并且具有自动存储期限，除非显式初始化，否则存储的值是不确定的.进一步使用它会导致未定义的行为.

引用C11，第§6.7.9章

如果未自动初始化具有自动存储持续时间的对象，则其值为 不确定.[....]

• 对于第一种情况，您有一个char类型，如果在这种情况下，它可以具有陷阱表示形式，它将导致UB，否则它将是一个随机值. /p>

• 在第二种情况下，如果您使用的是char *类型，则指针保存不确定的值，该值在程序上下文中为 invalid ，因此尝试取消引用该指针肯定会调用未定义的行为.

Below is the code where I am trying to print default values/values of char variable & pointer. But Can not able to see it on console.Is it has default values or just not able to read ASCII range.?

#include <stdio.h>
int main()
{
    char c, *cp;

    printf("\nValue of char c:%c\n", c);
    printf("\nValue of char ptr:%c\n", *cp);

    return 0;
}

解决方案

Surprise!! there is no "default" value here, you're venturing into undefined behavior.

To elaborate, if a variable is local scoped and automatic storage duration, unless initialized explicitly, the stored value is indeterminate. Using that further, causes undefined behavior.

Quoting C11, chapter §6.7.9

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.[....]

• For the first case, where you have a char type, if in your case, which can have a trap representation, it'll lead to UB, otherwise, it'll be a random value.

• For the second case, where you have a char * type, the pointer holds indeterminate value, which is invalid in context of your program, so attempt to dereference the pointer surely invokes undefined behavior.

这篇关于char指针或char变量的默认值是什么的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！