C++ “const”错误之前的预期主表达式
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expected primary-expression before ‘const’ errors
提问by user3330980
Please help. I am getting many errors.
请帮忙。我收到很多错误。
sub2.cpp: In function ‘int main()': sub2.cpp:11:14: error: invalid conversion from ‘const char*' to ‘char' [-fpermissive] sub2.cpp:12:14: error: invalid conversion from ‘const char*' to ‘char' [-fpermissive] sub2.cpp:16:17: error: expected primary-expression before ‘const' sub2.cpp:16:36: error: expected primary-expression before ‘const' sub2.cpp:11:6: warning: unused variable ‘outer' [-Wunused-variable] sub2.cpp:12:6: warning: unused variable ‘inner' [-Wunused-variable] make: *[sub2] Error 1
sub2.cpp：在函数“int main()”中：sub2.cpp:11:14：错误：从“const char*”到“char”的无效转换[-fpermissive] sub2.cpp:12:14：错误：无效从 'const char*' 到 'char' 的转换 [-fpermissive] sub2.cpp:16:17：错误：'const' 之前的预期主表达式 sub2.cpp:16:36：错误：'const' 之前的预期主表达式' sub2.cpp:11:6: 警告：未使用的变量 'outer' [-Wunused-variable] sub2.cpp:12:6: 警告：未使用的变量 'inner' [-Wunused-variable] 制作：*[sub2] 错误1
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
char *Subtract(const char *outer, const char *inner);
int main()
{
char outer = "Bookkepper";
char inner = "keep";
char *word = new char[50];
word = Subtract(const char &outer, const char &inner);
cout << word << endl;
return 0;
}
char *Subtract(const char *outer, const char *inner)
{
int olen = strlen(outer);
int first_occ_idx = -1;
for(int i=0; i < olen; i++){
if(strncmp(outer+i, inner,strlen(inner)) == 0){
first_occ_idx = i;
}
}
if(first_occ_idx == -1){
return NULL;
}
int ilen = strlen(inner);
int xx = olen - ilen;
char *newstr = new char[xx];
int idx = 0;
for(int i=0; i < first_occ_idx; i++){
newstr[idx++] = outer[i];
}
for(int i=first_occ_idx+ilen; i < olen; i++){
newstr[idx++] = outer[i];
}
newstr[idx] = 'const char *outer = "Bookkeeper"; // Note also spelling
';
return newstr;
}
采纳答案by paxdiablo
In C++, string literals like "Bookkepper"(sic)are constcharacter pointers,it's a little stricter than in C. So it should be:
在 C++ 中，像"Bookkepper"(sic)这样的字符串文字是const字符指针，它比 C 中的要严格一些。所以它应该是：
char outer = "Bookkepper";
rather than:
而不是：
word = Subtract(const char &outer, const char &inner);
In addition, you don't include types when callinga function, so:
此外，调用函数时不包括类型，因此：
word = Subtract(outer, inner);
would be better as:
会更好：
delete[] word;
Separately (and these are style suggestions only), the correct type for things that represent sizes(such as number of characters in a string) is size_trather than int.
另外（这些只是样式建议），表示大小（例如字符串中的字符数）的事物的正确类型是size_t而不是int.
And it's usually considered good form to clean up all your dynamic memory explicitly so, before returning from main(), you could put:
并且通常认为明确清理所有动态内存的好形式，因此，在从 返回之前main()，您可以输入：
