解释出现 "iso c++ forbids converting a string constant to ‘char*’" 警告的原因 在ISO C++标准中,字符串常量(如"Hello, World!")是存储在只读内存中的,其类型是const char[]。尝试将这些字符串常量赋值给char*类型的指针会导致类型不匹配的问题,因为char*意味着指向可修改字符数据的指针。这种转换是不安全...
string +2 parina 14 years ago 6 Comments (6) Write comment? daftcoder 14 years ago, # | 0 You should better to post your code on one of the services like codepad.org and have just one post, containing useful methods, e.g.: String converting...(string to char *) Sorting...
这个警告的原因是,C++ 标准禁止将字符串常量转换为 char* 类型,因为字符串常量是不可修改的,而 char* 指向的是一个可以修改内容的字符数组。这种转换会导致潜在的错误或不安全的操作,因为通过 char* 修改字符串常量的内容是未定义行为。 比如如下代码: char*text="Hello world";// 错误,会导致警告 具体原因: ...
char* p ="abc"; // valid in C, invalid in C++ 会跳出警告:warning: ISO C++ forbids converting a string constant to 'char*' [-Wwrite-strings] 改成下面会通过warning char* p = (char*)"abc"; // OK 或者改成下面: charconst*p ="abc"; // OK 原因解析: 我们在学习c或者c++的时候都知...
message() { ifstream inFile; string messageIn[2000]; char messageChar[2000]; int messageInt[2000]; inFile.open("message.dat", ios::in); if(inFile.fail()) { cout << "File did not open!"; Sleep(2000); exit(1); } else for ...
You don't need to use raw char arrays at all if you're already using strings. No need to use strcpy, just copy the std::strings by assignment. so I can search for the instruction within the string You want to search for a string within another string? You can also do that. ...
warning: ISO C++ forbids converting a string constant to 'char*' [-Wwrite-strings] 解决办法:采用第2或3种字符串赋值方式 intmain(intargc,char*argv[]) {charstr[] ="";//先把C++中的string常量复制给C语言形式的字符串变量,再将str赋值给char*形式的C语言字符串。argv[2] =str; ...
会跳出警告:warning: ISO C++ forbids converting a string constant to 'char*' [-Wwrite-strings] 改成下面会通过warning char* p = (char*)"abc"; // OK 1. 或者改成下面 char const *p = "abc"; // OK 1. 原因解析: 学习c或者c++的时候都知道,如果在赋值操作的时候,等号两边的变量类型不一样...
Hi, I got an error saying that my code below for the Hit, Miss, Total Miss, and Bullseye is currently a string and needs to be "of data type char" and I was wondering if someone could help function[zone,Points]=RQ18_18(X,Y) ...
一、错误代码展示 函数定义: voidreadImage(char*inputPath); 函数使用: readImage("C:\\xxxx\\girl.jpg"); 二、原因分析 在上面的方法中,方法的参数需要我们传递一个指针类型的字符。而我们在使用该方法的时候传递的确实一个常量。会导致常量强转为指针,因为会报这么一个警告。这个警告在有些编译器上就直接...