这两段代码是相同的吗？
char ch = 'a';
printf("%d", ch);
它会打印出一个垃圾值吗？
我对此很困惑。
printf("%d", '\0');
这段代码会打印出0还是垃圾值？因为当我执行这段代码时
printf("%d", sizeof('\n'));
它打印出4。为什么sizeof('\n')是4个字节呢？在C++中相同的东西打印1个字节。那是因为什么呢？
所以这里是主要问题
在C语言中，printf("%d", '\0')是否应该打印0?
而在C++中，printf("%d", '\0')是否应该打印垃圾值？
%d打印一个整数：它将打印出您字符的ASCII表示。你需要的是%c：
printf("%c", ch);
printf("%d", '\0'); 的输出结果是字符 '\0' 的 ASCII 码，即 0（通过对 0 进行转义，告诉编译器使用 ASCII 值 0）。
printf("%d", sizeof('\n')); 的输出结果是 4，因为在 C 语言中，字符字面量被视为 int 而不是 char。
这应该输出字符的ASCII值，因为%d是整数的转义序列。所以在打印时，作为printf参数给定的值将被视为整数。
char ch = 'a';
printf("%d", ch);
printf("%d", '\0');，其中空字符被解释为整数0。sizeof('\n')的结果是4，因为在C语言中，这种字符的表示法代表相应的ASCII整数。因此，'\n'作为整数与10相同。char类型在表达式中会被提升为int类型。如果你深思熟虑，这基本上解释了每一个问题。'\n'或'a'的类型为int（因此sizeof '\n' == sizeof (int)），而在C ++中它们的类型为char。printf("%d", '\0');应该只是打印0；表达式'\0'的类型为int，其值为0。printf("%d", ch);应该打印ch的整数编码（对于ASCII，'a' == 97）。#include <stdio.h>
#include <stdlib.h>
int func(char a, char b, char c) /* demonstration that char on stack is promoted to int !!!
note: this promotion is NOT integer promotion, but promotion during handling of the stack. don't confuse the two */
{
const char *p = &a;
printf("a=%d\n"
"b=%d\n"
"c=%d\n", *p, p[-(int)sizeof(int)], p[-(int)sizeof(int) * 2]); // don't do this. might probably work on x86 with gcc (but again: don't do this)
}
int main(void)
{
func(1, 2, 3);
//printf with %d treats its argument as int (argument must be int or smaller -> works because of conversion to int when on stack -- see demo above)
printf("%d, %d, %d\n", (long long) 1, 2, 3); // don't do this! Argument must be int or smaller type (like char... which is converted to int when on the stack -- see above)
// backslash followed by number is a oct VALUE
printf("%d\n", '\377'); /* prints -1 -> IF char is signed char: char literal has all bits set and is thus value -1.
-> char literal is then integer promoted to int. (this promotion has nothing to do with the stack. don't confuse the two!!!) */
/* prints 255 -> IF char is unsigned char: char literal has all bits set and is thus value 255.
-> char literal is then integer promoted to int */
// backslash followed by x is a hex VALUE
printf("%d\n", '\xff'); /* prints -1 -> IF char is signed char: char literal has all bits set and is thus value -1.
-> char literal is then integer promoted to int */
/* prints 255 -> IF char is unsigned char: char literal has all bits set and is thus value 255.
-> char literal is then integer promoted to int */
printf("%d\n", 255); // prints 255
printf("%d\n", (char)255); // prints -1 -> 255 is cast to char where it is -1
printf("%d\n", '\n'); // prints 10 -> Ascii newline has VALUE 10. The char 10 is integer promoted to int 10
printf("%d\n", sizeof('\n')); // prints 4 -> Ascii newline is char, but integer promoted to int. And sizeof(int) is 4 (on many architectures)
printf("%d\n", sizeof((char)'\n')); // prints 1 -> Switch off integer promotion via cast!
return 0;
}
是的，除非你很幸运，否则它会打印垃圾。
非常重要。
printf/sprintf/fprintf参数的类型必须与相关格式类型字符匹配。
如果类型不匹配并且编译通过，则结果非常未定义。
许多较新的编译器知道printf并发出警告，如果类型不匹配。如果您收到这些警告，请修复它们。
如果您想将变量函数的参数类型转换，必须提供强制转换（即显式转换），因为编译器无法确定需要执行转换（就像具有类型化参数的函数原型一样）。
printf("%d\n", (int) ch)
printf("%d", (int) sizeof('\n'))
printf("%d", ch);“打印出一些垃圾值”，我猜这是因为他想要打印出A，而不是它的ASCII值... - peoro%d打印字符的结果感到好奇，而%c则完全无关。 - orlp