UNIX上的C程序转储核心

Question

I am a new learner of C language.

Below program run well on Windows but when i compile with gcc on solaris, this is dumping core

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

void main()
{
char *name;
name="James Bond";
int i=0;
sprintf(name,"%s/%d",name,i);
printf("String is %s",name);
}

Please suggest

Answer 1

You cannot modify a string literal like that, it's undefined according to the standard. You're trying to overwrite that string literal with other data (the sprintf ).

Many implementations will place them in read-only memory, causing a core dump - they're the good ones. The bad ones will continue on as if everything's okay, which it usually isn't.

You could try the following:

#include <stdio.h>

int main (void) {
    char *name;
    char name2[100];  // make sure plenty of space.
    name = "James Bond";
    int i = 0;
    sprintf (name2, "%s/%d", name, i);
    printf ("String is %s\n", name2);
    return 0;
}

Most questions of this type have code like:

name = "Bob";
*name = 'J';   // to try and make "Job"

but it's just as undefined to write to string literals using sprintf as well.

---

Based on comments, you want to be able tocombine a path and file spec. You could do this as something like:

char *path = "/tmp/";
char *file = "xyz.txt"
char fullpath = malloc (strlen (path) + strlen (file) + 1);
if (fullpath == NULL)
    // error and exit condition
strcpy (fullpath, path);
strcat (fullpath, file);
// use fullpath for your nefarious purposes :-)
free (fullpath);

That's one way to do it, there are others.

Answer 2

char *name;
name="James Bond";   // name is pointing into read-only memory
int i=0;
sprintf(name,"%s/%d",name,i); // trying to write to read-only memory
printf("String is %s",name);

instead use a buffer

char name[32] = "James Bond";
...

Answer 3

The correct way to define and initialize a constan string in C is

char name[]="James Bond";

Your code may be like this:

#include<stdio.h>
#include<stdlib.h>
void main()
{
    char name[] = "James Bond";
    int i = 0;

        printf("String is %s/%d", name,i);

}

Answer 4

您的“名称”字符串长度不足，无法容纳使用sprintf向其打印的字符集-您需要为所有字符分配足够的缓冲区。

Answer 5

您需要使用malloc / calloc或将其定义为固定长度的变量来为name指针分配内存： char name[50] （例如）

Answer 6

On Linux and systems with the GNU Libc, you can also code using asprintf

char* str = NULL; // pointer should be initialized to NULL
asprintf (&str, "someformat %d", 20);

When using GTK, you could also call g_strdup_printf

You should understand the difference and similarities between arrays and pointers in C. Many many books or lectures explain that in detail.

Regards.