新的c ++有一些问题 [英] new at c++ have some questions
问题描述
我正在尝试学习C ++并且有一些问题。我得到一个
分段错误，无论是什么，当我尝试这个简单的程序时。
基本上该程序假设打印五 ;。我一直在printf上阅读
，我想我发现了％s符号的错误。它说％s是
假设打印字符串但它不起作用！
谢谢你！
＃包括< stdio.h>
#define is =
#define start main（）
start
{
int x是5;
printf（" x的值是％s \ n"， x）;
}
wintaki写道：
我正在尝试学习C ++并有一些问题。当我尝试这个简单的
程序时，我得到了一个
分段错误，无论如何。
基本上是程序假设打印五。我一直在读取printf上的b $ b，我认为我发现了％s符号的错误。
它说％s假设打印字符串但它不起作用！
谢谢你！
#include< stdio.h>
#define is =
#define start main（）
如果你想编写C ++，首先要抛弃那种宏。他们
不会给语言添加任何清晰度，他们只会让你更难理解。
start
宏解析，main（）不是C ++。 int main（）是。
{
int x是5;
printf（ x的值是％s \ n，x）;
}
现在，当然你得到一个段错误。 ％s应该打印一个
*字符串*。
你提供的是* int *。阅读printf的文档
（完全！）。
干杯，
-
IR
wintaki写道：
我想尝试学习C ++并提出一些问题。当我尝试这个简单的程序时，我得到了一个
分段错误，无论是什么。
这意味着你的程序试图访问不属于
的内存。
基本上该程序假设打印五。我一直在printf上阅读
，我想我发现了％s符号的错误。它说％s是
假设打印字符串但它不起作用！
这是你程序中的一个错误。 ％s确实用于打印字符串。问题
是你不提供字符串，而是提供int。由于printf没有输入
安全，所以它没有注意到这个问题。相反，它只是尝试从实际存储int的位置读取字符串
。
#include< stdio.h>
#define is =
#define start main（）
那些应该是什么除了混淆代码之外还好吗？
start
{
int x是5;
printf（" x的值是％s \ n"，x）;
}
Rolf Magnus写道：
wintaki写道：
我正在尝试学习C ++并有一些问题。当我尝试这个简单的程序时，我得到了一个
分段错误，无论是什么。
这意味着你的程序试图访问不属于
的内存。
这肯定不是我想做的。我不想访问
别人的记忆。我的目标是从5开始并打印
字符串版本。据我所知，字符串是一个文本值，并且
" 5"是一个int值。
>
基本上该程序打算打印五。我一直在printf上阅读
，我想我发现了％s符号的错误。它说％s是
假设打印字符串但它不起作用！
这是你程序中的一个错误。 ％s确实用于打印字符串。问题
是你不提供字符串，而是提供int。由于printf没有输入
安全，所以它没有注意到这个问题。相反，它只是尝试从实际存储int的位置读取字符串
。
#include< stdio.h>
#define is =
#define start main（）
那些应该是什么除了混淆代码之外还好吗？
ok我会停止那些，我只是测试它，因为我是新来的
#define并认为能够自定义是很好的C ++到我的
喜欢。但我明白你的观点。
然后新程序是
int main（）
{
int x = 5;
printf（值为％s \ n，x）;
}
但我仍有同样的问题。说的是我不能用％s使用％s
？但是，我如何获得文本值？我想
写一个简单的程序，取x值并用文字打印。
i认为％s将值转换为文本？
感谢
Hi,
I am trying to learn C++ and have some questions. I get a
"Segmentation fault", whatever that is, when I try this simple program.
Basically the program is suppose to print "five". I have been reading
on printf and I think I found a bug with the %s symbol. it says %s is
suppose to print the string but it does not work!
Thanks be to you!
#include <stdio.h>
#define is =
#define start main()
start
{
int x is 5;
printf("the value of x is %s\n", x);
}
wintaki wrote:
Hi,
I am trying to learn C++ and have some questions. I get a
"Segmentation fault", whatever that is, when I try this simple
program.
Basically the program is suppose to print "five". I have been
reading on printf and I think I found a bug with the %s symbol.
it says %s is suppose to print the string but it does not work!
Thanks be to you!
#include <stdio.h>
#define is =
#define start main()If you want to write C++, first ditch that kind of macros. They
don''t add any clarity to the language, they just make it harder to
understand.
startMacros resolved, main() is not C++. int main() is.
{
int x is 5;
printf("the value of x is %s\n", x);
}Now, of course you get a segfault. %s is supposed to print a
*string*.
What you provide it is an *int*. Read printf''s documentation
(completely!).
Cheers,
--
IR
wintaki wrote:
Hi,
I am trying to learn C++ and have some questions. I get a
"Segmentation fault", whatever that is, when I try this simple program.It means that your program attempts to access memory that doesn''t belong to
it.
Basically the program is suppose to print "five". I have been reading
on printf and I think I found a bug with the %s symbol. it says %s is
suppose to print the string but it does not work!It''s a bug in your program. %s is indeed used to print strings. The problem
is that you don''t provide a string, but an int. Since printf isn''t type
safe, it won''t notice the problem. Instead, it just tries to read a string
from where actually an int is stored.
#include <stdio.h>
#define is =
#define start main()What are those supposed to be good for, other than obfuscate the code?
start
{
int x is 5;
printf("the value of x is %s\n", x);
}
Rolf Magnus wrote:wintaki wrote:
Hi,
I am trying to learn C++ and have some questions. I get a
"Segmentation fault", whatever that is, when I try this simple program.
It means that your program attempts to access memory that doesn''t belong to
it.this is certainly not what i want to do. i do not wish to access
someone else''s memory. my goal is to start with 5 and have it print
the string version. as I understand it a string is a text value, and a
"5" is an int value.
>Basically the program is suppose to print "five". I have been reading
on printf and I think I found a bug with the %s symbol. it says %s is
suppose to print the string but it does not work!
It''s a bug in your program. %s is indeed used to print strings. The problem
is that you don''t provide a string, but an int. Since printf isn''t type
safe, it won''t notice the problem. Instead, it just tries to read a string
from where actually an int is stored.
#include <stdio.h>
#define is =
#define start main()
What are those supposed to be good for, other than obfuscate the code?
ok I will stop with those, I was just testing it out since I am new to
#define and thought it was nice to be able to "customize" C++ to my
liking. but I see your point.
But then the new program is
int main()
{
int x = 5;
printf("the value is %s\n", x);
}
but i still have the same problem. are saying is that I can not use %s
with an int value? but then how do I get a text value? I wanted to
write a simple program that took the x value and printed it in words.
i thought %s converted the value to text?
thanks
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