memcpy error : Segmentation fault (core dumped)

问题

I'm trying to copy one string to another in c using memcpy with the following code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


typedef struct fullName
{
    char* firstName;
    char* lastName;
} NAME;

int main()
{              
    NAME myName, hisName;

    myName.firstName =  "aaaaaaa";
    hisName.firstName = "bbbbbb";

    memcpy(myName.firstName, hisName.firstName, sizeof(hisName.firstName));

    printf("myName.firstName = %s\n", myName.firstName);
    printf("hisName.firstName = %s\n", hisName.firstName);
}

and it gives Segmentation fault (core dumped) Error after running the program I tried to declare firstName and lastName as pointer to an array of chars rather than pointer to char but the error persists . what am I missing ?! plz help...

FYI .. I'm working on Ubuntu 14 and I'm using gcc (Ubuntu 4.8.2-19ubuntu1)...

回答1:

You are assigning myName.firstName and hisName.firstName with pointers to string literals. String literals cannot be modified, which is what's causing your error.

To achieve what you want, you can either declare firstName as a char array, or allocate memory to it (as a char pointer).

Array method:

typedef struct fullName
{
        char firstName[256]; // a sufficiently large number
        char lastName[256];
} NAME;

int main()
{              
        NAME myName, hisName;
        strcpy(myName.firstName, "aaaaaaa"); // You can't assign a string directly
        strcpy(hisName.firstName, "bbbbbb");
        memcpy(myName.firstName, hisName.firstName, sizeof(hisName.firstName));
        printf("myName.firstName = %s\n", myName.firstName);
        printf("hisName.firstName = %s\n", hisName.firstName);

}

Allocation method:

typedef struct fullName
{
        char* firstName;
        char* lastName;
} NAME;

int main()
{              
        NAME myName, hisName;
        size_t buffersize = 256; // a sufficiently large number
        myName.firstName = malloc(buffersize);
        hisName.firstName = malloc(buffersize); // same for lastName

        strcpy(myName.firstName, "aaaaaaa");
        strcpy(hisName.firstName, "bbbbbb");
        memcpy(myName.firstName, hisName.firstName, buffersize); // You can't use sizeof() as this will give the size of a pointer
        printf("myName.firstName = %s\n", myName.firstName);
        printf("hisName.firstName = %s\n", hisName.firstName);

}

回答2:

In these statements

myName.firstName =  "aaaaaaa";
hisName.firstName = "bbbbbb";

you initialized pointers with addresses of string literals.

In next statement

memcpy(myName.firstName, hisName.firstName, sizeof(hisName.firstName));

you try to modify one of the string literals.

According to the C Standard (6.4.5 String literals)

7 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

Also this expression

sizeof(hisName.firstName)

returns size of the pointer itself. It is not the same as the size of the corresponding string literal.

The valid program could look the following way

#include <stdio.h>
#include <string.h>


typedef struct fullName
{
    char firstName[8];
    char lastName[8];
} NAME;

int main()
{              
    NAME myName = { "aaaaaaa" };
    NAME hisName = { "bbbbbb" };

    memcpy( myName.firstName, hisName.firstName, sizeof( hisName.firstName ) );

    printf( "myName.firstName = %s\n", myName.firstName );
    printf( "hisName.firstName = %s\n", hisName.firstName );
}

回答3:

This did not crash in MSVC, it produced the output:

myName.firstName = bbbbaaa
hisName.firstName = bbbbbb

Notice that only 4 chars from hisName have been copied to myName. That is because

sizeof(hisName.firstName)

is 4 on your platform, the size of the pointer. But manipulating such strings is Undefined Behaviour anyway.

来源：https://stackoverflow.com/questions/29523680/memcpy-error-segmentation-fault-core-dumped

标签

pointers

memcpy