C char指针和函数参数传递

Question

The two function getData and getData2 all can get right answers, are they valid?

 #include<stdio.h>
 #include<stdlib.h>
 void getData(const char** data) {
     if(data == NULL) {
         printf("NULL\n");
     }
     *data = "error";
 }
 const char* getData2() {
     const char*p = "hello";
     return p;
 }
 int main(){
     const char *p = NULL;
     getData(&p);
     printf("data:%s\n",p);
     printf("data2:%s\n",getData2());
 }

Answer 1

char* p = "hello";

This is not allowed since C++11, but earlier versions allow it.

The string literal "hello" is stored in read-only memory that can't be modified, but a pointer to a non-const char has the ability to let the memory be modified, which will crash during runtime when pointing at a string literal.

A modern compiler won't accept such a conversion from const char* to char* .

Same with the char** data parameter and *data = "error"; assignment. data should be type as const char** to make the assignment legal.