将数组复制到向量中
我编写了一个小程序:
void showrecord()
{
char *a[]={ "O_BILLABLE_ACCOUNT","O_CUSTOMER_TYPE_INDICATOR",
"O_A_PARTY_MSISDN_ID","O_A_PARTY_EQUIPMENT_NUMBER",
"O_A_PARTY_IMSI","O_A_PARTY_LOCATION_INFO_CELL_ID",
...
};
vector<std::string> fields(a,a+75);
cout<<"done!!!"<<endl;
}
int main()
{
showrecord();
}
我有字符串文字数组,我希望将它们复制到向量中。 我没有找到任何其他简单的方法来做到这一点:(。或者如果有任何直接的方法来初始化向量而不使用数组,那将非常有帮助。 这是我在 UNIX 上运行可执行文件后转储核心。 它给了我一个警告,例如:
Warning 829: "test.cpp", line 12
# Implicit conversion of string literal to 'char *' is deprecated.
D_TYPE","O_VARCHAR_5","O_VARCHAR_6","O_VARCHAR_7","O_VARCHAR_8","O_VARCHAR_9"};
但是相同的代码在 Windows 上运行良好,没有任何问题。 我在 HPUX 上使用编译器 aCC。
请帮忙! 编辑 下面是转储的堆栈跟踪。
(gdb) where
#0 0x6800ad94 in strlen+0xc () from /usr/lib/libc.2
#1 0xabc0 in std::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string<char,std::char_traits<char>,std::allocator<char>>+0x20 ()
#2 0xae9c in std<char const **,std::basic_string<char,std::char_traits<char>,std::allocator<char>> *,std::allocator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>>::uninitialized_copy+0x60 ()
#3 0x9ccc in _C_init_aux__Q2_3std6vectorXTQ2_3std12basic_stringXTcTQ2_3std11char_traitsXTc_TQ2_3std9allocatorXTc__TQ2_3std9allocatorXTQ2_3std12basic_stringXTcTQ2_3std11char_traitsXTc_TQ2_3std9allocatorXTc____XTPPCc_FPPCcT118_RW_is_not_integer+0x2d8
()
#4 0x9624 in showrecord () at test.cpp:13
#5 0xdbd8 in main () at test.cpp:21
I have written a small program:
void showrecord()
{
char *a[]={ "O_BILLABLE_ACCOUNT","O_CUSTOMER_TYPE_INDICATOR",
"O_A_PARTY_MSISDN_ID","O_A_PARTY_EQUIPMENT_NUMBER",
"O_A_PARTY_IMSI","O_A_PARTY_LOCATION_INFO_CELL_ID",
...
};
vector<std::string> fields(a,a+75);
cout<<"done!!!"<<endl;
}
int main()
{
showrecord();
}
I have array of string literals and i want them to be copied into a vector.
I did not find any other easy way to do it :(.Or if there is any direct way to initialize the vector without using the array ,that would be very much helpful.
This is dumping the core after i run the executable on unix.
It gives me a warning though like :
Warning 829: "test.cpp", line 12
# Implicit conversion of string literal to 'char *' is deprecated.
D_TYPE","O_VARCHAR_5","O_VARCHAR_6","O_VARCHAR_7","O_VARCHAR_8","O_VARCHAR_9"};
But the same code is running fine on windows without any problem.
I am using the compiler aCC on HPUX.
Please help!
EDIT
below is teh stacktrace of the dump.
(gdb) where
#0 0x6800ad94 in strlen+0xc () from /usr/lib/libc.2
#1 0xabc0 in std::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string<char,std::char_traits<char>,std::allocator<char>>+0x20 ()
#2 0xae9c in std<char const **,std::basic_string<char,std::char_traits<char>,std::allocator<char>> *,std::allocator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>>::uninitialized_copy+0x60 ()
#3 0x9ccc in _C_init_aux__Q2_3std6vectorXTQ2_3std12basic_stringXTcTQ2_3std11char_traitsXTc_TQ2_3std9allocatorXTc__TQ2_3std9allocatorXTQ2_3std12basic_stringXTcTQ2_3std11char_traitsXTc_TQ2_3std9allocatorXTc____XTPPCc_FPPCcT118_RW_is_not_integer+0x2d8
()
#4 0x9624 in showrecord () at test.cpp:13
#5 0xdbd8 in main () at test.cpp:21
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评论(3)
为什么是75?
更改
为
没有可以说是“更好”的方法来初始化 C++03 的向量,但对于 C++0x,您可以使用更方便的语法,无需使用 C 数组:
Why 75?
Change
to
There's no arguably 'better' way to initialize your vector for C++03, but for C++0x you have access to a more convenient syntax, dispensing with the C array:
尝试使用
const char* a[]
而不是char* a[]
。字符串文字的类型是const char*
,而不是char*
,因此您会收到警告。Try
const char* a[]
instead ofchar* a[]
. String literals are of typeconst char*
, notchar*
, and hence you get the warning.这是可能的解决方案,IMO 更通用一些 - 使用可重用的函数,可处理任何大小的字符串数组:
Here is possible solution which IMO is a little bit more general - uses reusable function that works with string arrays of any size: