https://stackoverflow.com › questions › 58043897

This is because your expression newItemIDPointer + i is a pointer to the character at offset i in the string, not the value (character) at that location. You need to dereference the pointer to get the value, eg: *(newItemIDPointer + i) Or a more obvious way to do it is: newItemIDPointer[i]

https://stackoverflow.com › questions › 55751920

The problem you're having is that it's illegal to drop const qualifiers (with the exception of the infamous const_cast or equivalent C-style cast). Since you're only reading from sir, you can fix this by making sir be const char* instead, which doesn't violate const: class student {.

https://cs50.stackexchange.com › questions › 38804

This function counts lenght of a string entered at the command line argument. I get an error :" incompatible pointer types passing 'string' (aka 'char *') to parameter of type 'string *' (aka 'char **') ". Could You help me on this.

https://learn.microsoft.com › en-us › cpp › error-messages › compiler-errors-1 › compiler-error...

The compiler generates C2440 when it can't convert from one type to another, either implicitly or by using the specified cast or conversion operator. There are many ways to generate this error. We've listed some common ones in the Examples section.

https://wenku.csdn.net › answer › 9c700c977c724831a6813566ff3f607f

这个错误通常是因为你在函数调用中传递了一个 int 类型的参数，但函数所期望的参数类型是 unsigned char *。 你需要检查一下函数原型和调用代码，确认参数类型是否匹配。 如果你确定参数类型是正确的，你可以尝试将参数强制转换为 unsigned char * 类型，例如： int value = 42; function((unsigned char *)&value); 注意，在进行强制类型转换时需要谨慎，确保不会出现意外的类型错误。 相关问题. (65): error: #167: argument of type "int" is incompatible with parameter of type "u8 *"

https://cplusplus.com › forum › windows › 271990

argument of type "wchar_t *" is incompatible with parameter of type "LPWSTR" What is the best way to resolve this error? I can declare the variable as a string pointer.

https://www.codeproject.com › ... › 5319780 › Cplusplus-argument-of-type-is-incompatible-with-pa

1 solution. Solution 1. C++ now enforces the use of the const qualifier on constant values> Change your function to: int diap(const char * s, int low, int high) Also, why are you inputting double types when all your comparisons and values are int types in the rest of the program?

https://cboard.cprogramming.com › cplusplus-programming › 179265-problem-argument-type...

The reason why this is a red flag is that what a char* parameter points to could be modified, whereas a string literally cannot be safely modified. In your case, the simple solution is to change the parameter to be a const char* since the function really doesn't modify what the pointer points to.

https://www.reddit.com › ... › wbxdhy › const_char_is_incompatible_with_parameter_of_type

The error you’re getting is because arrays in C & C++ implicitly convert to pointers to the first element of the array. Individual chars are in single quotes. In your example, changing the double quotes to single quotes should fix the problem