无法打印char！ [英] Can't print char!
问题描述
我想打印字符''d''：
main（）{
char g [4];
g [0] =''a'';
g [1] =''b'';
g [2] =''c '';
g [3] =''d'';
char * pp;
pp =& g [3];
printf（"％s"，* pp）;
getchar（）;
}
但如果我改变的话，它唯一可以打印：
printf（"％s"，* pp） ;
进入：
printf（"％s"，pp）;
但是当我用int做同样的事情时我必须使用第一个版本的
printf：
main（）{
int j = 50;
int * q;
q =& j;
printf（"％d"，* q）;
getchar（）;
}
为什么它与整数相反？
JS
与％s printf中打印一个字符串，由
第一个字符的一个指针来表示。它只是运气，它打印出一个字符，如果
以下地址的值为非零，它会打印出来
垃圾（ - >一个字符串是null终止）。
要打印一个整数，你需要取消引用指针，因为printf
需要的是值，而不是地址。
simon
在文章< d1 ********** @ news.net.uni-c.dk> ;, JS< sf****@asdas.com>写道：
：我想打印字符''d''：
：char g [4];
：g [3] =''d'';
：pp =& g [3];
：printf（" ;％s"，* pp）;
* pp是单个字符，而不是字符串。您应该使用％c
格式进行打印 - 否则 - 'd''对应的值 -
将用作指向字符串打印，结果令人讨厌。
你也犯了一个错误，使用％s格式来处理
a字符数组，它不是NULL终止。
printf（"％s"，pp）;将要从字符串的末尾开始运行
这是一件坏事。
-
如果你喜欢VT-52，那就太开心了的。
JS< sf **** @ asdas.com>这样说：
main（）{
在C99中无效。 main（）返回int。
char g [4];
g [0] =''a'';
g [1] =''b'';
g [2] =''c'';
g [3] =''d'';
char * pp;
pp =& g [3];
pp指向g的最后一个元素，即一个字符。
printf（"％s"，* pp）;
你可能意味着
printf（"％c \ nn"，* pp）;
的getchar（）;
在C89中不允许省略main（）的退货声明。
printf（"％s"，pp）;
非常糟糕。 ％s格式说明符需要以null结尾的
字符数组，而pp则不需要。未定义的行为结果。
为什么它与整数相反？
因为你使用了不正确的格式说明符，如上所述。
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
I would like to print char ''d'':
main(){
char g[4];
g[0] = ''a'';
g[1] = ''b'';
g[2] = ''c'';
g[3] = ''d'';
char *pp;
pp = &g[3];
printf("%s", *pp);
getchar();
}
But its only possible to get d printed if I change:
printf("%s", *pp);
into:
printf("%s", pp);
But when I do the same thing with an int I have to use the first version of
the printf:
main(){
int j = 50;
int *q;
q = &j;
printf("%d", *q);
getchar();
}
Why is it the other way around with integers??
JS
with %s in printf you print a string, represented by a pointer of the
first char. it is just luck that it prints out one char, if the
following address would have a value of nonzero, it would print out
garbage (-> a string is null terminated).
to print an integer, you need to dereference the pointer, because printf
needs the value, not the address.
simon
In article <d1**********@news.net.uni-c.dk>, JS <sf****@asdas.com> wrote:
:I would like to print char ''d'':
: char g[4];
: g[3] = ''d'';
: pp = &g[3];
: printf("%s", *pp);
*pp is a single character, not a string. You should be using a %c
format to print it -- otherwise the -value- that ''d'' corresponds to
will be used as a pointer to the string to print, with nasty results.
You also are making a mistake by using a %s format to deal with
a character array that is not NULL terminated.
printf("%s", pp); is going to run off the end of the string
which is a Bad Thing.
--
Feep if you love VT-52''s.
JS <sf****@asdas.com> spoke thus:
main(){
Invalid in C99. main() returns int.
char g[4];
g[0] = ''a'';
g[1] = ''b'';
g[2] = ''c'';
g[3] = ''d''; char *pp;
pp = &g[3];
pp points to the last element of g, which is a character.
printf("%s", *pp);
You probably meant
printf( "%c\n", *pp );
getchar();
Omitting the return statement of main() is not permitted in C89.
printf("%s", pp);
Very bad. The %s format specifier requires a null-terminated
array of characters, which pp is not. Undefined behavior results.
Why is it the other way around with integers??
Because you used the incorrect format specifier, as above.
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
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