[英]difference between printf() and std::cout with respect to pointers
I am new to pointers and i cant figure out one simple thing.我是指针的新手,我无法弄清楚一件简单的事情。
int main ()
{
char *str1="pointer";
printf("%p \n", str1);
cout << str1<<endl;
return 0;
}
The output is as follows :输出如下:
0000000000409001
pointer
Could someone please explain me the difference here.有人可以解释一下这里的区别吗。 why isnt cout printing the memory address ?
为什么不打印内存地址? how can i make cout print the address of str1?
如何让 cout 打印 str1 的地址?
The format specifier %p
prints a void *
, (untrue: so the char *
is implicitly converted to void *
) the char *
is converted to void *
before printing.格式说明符
%p
打印一个void *
,(不真实:所以char *
隐式转换为void *
) char *
void *
在打印之前转换为void *
。 (But this is actually undefined behavior, see comments. The correct way to do that would be printf("%p", (void *) str1);
) The corresponding C++ code would be std::cout << (void *) str1 << '\\n';
(但这实际上是未定义的行为,请参阅注释。正确的方法是
printf("%p", (void *) str1);
)相应的 C++ 代码将是std::cout << (void *) str1 << '\\n';
. .
The code std::cout << str1;
代码
std::cout << str1;
prints str1
as null terminated string.将
str1
打印为空终止字符串。 The corresponding C-code would be printf('%s', str1);
相应的 C 代码将是
printf('%s', str1);
A pointer is an address to a location in memory.指针是指向内存中某个位置的地址。
"pointer"
is a C-string in memory, 8 bytes for the letters and a terminating NULL byte. "pointer"
是内存中的一个 C 字符串,8 个字节的字母和一个终止的 NULL 字节。 str1
is a pointer to the byte of the first letter 'p'
. str1
是指向第一个字母'p'
的字节的指针。
printf("%p", str1)
prints the value of the pointer itself, that is the memory address (in this case 0000000000409001
). printf("%p", str1)
打印指针本身的值,即内存地址(在本例中为0000000000409001
)。
printf("%s", str1)
would print pointer
, the content of the C-string at location str1
. printf("%s", str1)
将打印pointer
,即位置str1
处的 C 字符串的内容。
cout << str1 << endl
also prints the content of the C-string. cout << str1 << endl
还会打印 C 字符串的内容。 This is the default behavior for pointer of type char*
because they are usually strings.这是
char*
类型指针的默认行为,因为它们通常是字符串。
cout << static_cast<void*>(str1) << endl
would print the address of the pointer again. cout << static_cast<void*>(str1) << endl
将再次打印指针的地址。
a char* is a pointer to the beginning of an array of characters. char* 是指向字符数组开头的指针。
cout
"recognizes" a char* and treats it like a string. cout
“识别”一个 char* 并将其视为字符串。
You are explicitly telling printf() to print out the decimal representation of a pointer address with the %p formatter.
您明确告诉 printf() 使用 %p 格式化程序打印出指针地址的十进制表示。
You are explicitly telling printf() to print out a representation of the pointer address with the %p formatter.
您明确告诉 printf() 使用 %p 格式化程序打印出指针地址的表示。
EDIT: edited for accuracy based on comment编辑:根据评论为准确性进行编辑
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