How to change the value of the string char * array = “abc” through a function?

1. Your variable array is of type const char* , not char* .
2. Because there is an implicit copying of the transferred data.

In order to write a C++ pointer to a string, you need:

1. Allocate memory from the heap.
2. Pass the pointer by reference, or pass the pointer to the pointer, "dereference" the pointer, and write data to it.

 #include <iostream> #include <cstring> #include <stdlib.h> using namespace std; void chchar(char** str) { strcpy(*str, "String data from `chchar`"); } int main(int argc, char** argv) { char* data = static_cast<char*>(malloc(sizeof(100))); strcpy(data, "Blablbla"); printf("Data before call `chchar`: %s\n", data); chchar(&data); printf("Data after call `chchar`: %s\n", data); free(data); system("pause"); } 

You can also pass a pointer to a string by reference:

 #include <iostream> #include <cstring> #include <stdlib.h> using namespace std; void chchar(char*& str) { strcpy(str, "String data from `chchar`"); } int main(int argc, char** argv) { char* data = static_cast<char*>(malloc(sizeof(100))); strcpy(data, "Blablbla"); printf("Data before call `chchar`: %s\n", data); chchar(data); printf("Data after call `chchar`: %s\n", data); free(data); system("pause"); } 

The result of both cases:

 Data before call `chchar`: Blablbla Data before call `chchar`: String data from `chchar` 

Example: https://ideone.com/1W85dS

In the function, you change the value of a pointer to a string, thus you do not change the external pointer variable, but change the pointer address, in the end it turns out that you just changed the address of a local variable.

To bring the change out, you need a pointer to a pointer, and change it.

 void chchar(char** a); int main() { char* array [] = {"sdf"}; chchar(array ); std::cout << "Hello, " << *array << "!\n"; } void chchar(char** a){ *a = {"asfafgag"}; }