can you tell me why11?

Code:

---

#include <stdio.h>
#include <string.h>
int main(int argc,char * argv[])
{int i;
char buf[4];
strcpy (buf,"aaaa");
printf("%d\n",i);
return 0;
}

---

Run-Time Check Failure #3-the variable i is being used without being defined
why ........

1st, you're trying to print the value of i, but you have not assigned a value to i. So there's no telling what's in the memory location containing i.

2nd, you're trying to copy a 4 char string into an array of 4 elements, you need 1 more element to hold the '\o' null terminating char.

[edit] Oh just noticed too... if you're trying to program in C++ you should be using cout << i instead of the printf() statement. I won't bother about the format of your #includes because that just may be because you're using an older compiler.

Code:

---

#include <stdio.h>
#include <string.h>
int main(int argc,char * argv[])
{int i=3;
char buf[5];
strcpy (buf,"aaaa");
printf("%d\n",i);
return 0;
}

---

like this ??..

Code:

---

#include <stdio.h>
main()
{ char buff[4];
 strcpy(buff,"aaaa");
 printf("%s",buff);
}

---

why these codes can run without mistake??

Because 5 chars "aaaa\0" is larger than 4 chars (char buff[4])

Adding to what Salem said:

A 'string literal' like "aaaa" is actually stored in memory like this:

aaaa\0

So, when you use strcpy() to try and copy that into an array with length=4, it won't work.

Quote:

---

Originally Posted by pacific

Code:

---

#include <stdio.h>
main()
{ char buff[4];
 strcpy(buff,"aaaa");
 printf("%s",buff);
}

---

why these codes can run without mistake??

---

Since you have written into memory where you weren't supposed to, the behavior is undefined. Undefined means that the program might or might not give the appearance of running as expected (maybe the memory you wrote to was not being used for anything else at that time). It might do anything. If you are lucky it will give a run-time error so that you know that something is wrong. If you are not lucky, the program will appear to run OK and will give expected output. Then, sometime in the future, this program, or a program that incorporates this code will cause user problems that you didn't forsee.

Nothing in the C language checks for memory accesses out of bounds. It is the programmer's responsibility.

This is extremely important, and it is an illustration that you can't prove a program is correct by testing it.

When I ran the following, I could see the consequences of out-of-range array access:

Code:

---

#include <stdio.h>
#include <string.h>
int main(void)
{
  int x=-1;
  char buff[4];
  int y=-1;

  printf("Before strcpy:\n");
  printf("x = %d\n", x);
  printf("y = %d\n\n", y);

  strcpy(buff,"aaaa");
  printf("buff: <%s>\n\n",buff);

  printf("After  strcpy:\n");
  printf("x = %d\n", x);
  printf("y = %d\n", y);

  return 0;
}

---

With my compilers I get this (no compiler error or warning messages; no run-time messages):

Quote:

---

Before strcpy:
x = -1
y = -1

buff: <aaaa>

After strcpy:
x = -256
y = -1

---

What do you get?

Regards,

Dave