练习在 printf 函数中使用指针时程序不断崩溃

Question

I'm new to programming. I'm attempting to run this extremely simple program as practice to integrate pointers into my code. When I run the program, it crashes on startup with a Windows has encountered an unexpected error pop-up. Absolutely no errors are given for the code itself, but the pop-up persists.

I'm running the code so far on Code::Blocks, and the program comes up with the stated pop-up from Windows on startup. I have also tested the program on codepad.org and the only error message that comes up has to do with the first line (the include statement). The error states: Segmentation fault .

#include <stdio.h>
int main() {
    // Setting a variable to store the value 15.
    unsigned short int random_number = 15;
    // Defining my pointer.
    unsigned short int *number_pointer;
    // Setting the pointer to hold the memory address of my first variable.
    number_pointer = random_number;
    // Attempting to print the value at the memory address stored in the pointer.
    printf("This code will now print a number: %i", *number_pointer);
    return 0;
}

I expect the output to be: This code will now print a number: 15. , however I didn't even get an output due to the program crashing.

Answer 1

Here

number_pointer = random_number;

number_pointer is of pointer type, and it should initialized with some valid address , not some value. In fact above statement causes compiler to warn you like

warning: incompatible integer to pointer conversion assigning to 'unsigned short *' from 'unsigned short'; take theaddress with & [-Wint-conversion]

but you seems ignored that. Never ignore compiler warnings. Always compile code with minimal warning flags like -Wall & read those warnings. for eg

gcc -Wall -Wextra -Wpedantic -Werror test.c  

The number_pointer needs to be point to address of random_number , so that you can de-reference it like *number_pointer .

It should be

number_pointer = &random_number; /* assign the address of random_number */

Answer 2

The comment

// Setting the pointer to hold the memory address of my first variable.

seems correct but the commented code is not:

number_pointer = random_number;

random_number is a number, number_pointer is a pointer.
Number and pointer are different types and the compiler should warn you that they are incompatible types.

Assigning a value to a variable of an incompatible type leads to data loss (when the variable's type uses less bytes than the value's type). But the main danger when data of incompatible types is assigned is the corruption of the data.

In your case, the value of number_pointer ( 15 , which is a number) is stored into the variable random_pointer . When the value of random_pointer is used, 15 is interpreted as a pointer, ie an address in memory and this is completely wrong.

The expression *number_pointer then tries to read the memory at that address and this leads to an access error that is penalized by the OS (which terminates your program).

The source of the problem is the missing & (the "address of" operator) in front of random_number in the assignment above. It should read:

number_pointer = &random_number;

This means the variable number_pointer (of type "pointer to an int") will store the address of variable random_number (of type "int"). They are compatible and the logic of the program is correct. Using the dereference operator ( * ) then successfully reads the integer value stored at the address stored in the number_ponter variable.