C Programming, Part 3: Common Gotchas

What common mistakes do C programmers make?

Memory mistakes

String constants are constant

char array[] = "Hi!"; // array contains a mutable copy 
strcpy(array, "OK");

char *ptr = "Can't change me"; // ptr points to some immutable memory
strcpy(ptr, "Will not work");

String literals are character arrays stored in the code segment of the program, which is immutable. Two string literals may share the same space in memory. An example follows:

char * str1 = "Brandon Chong is the best TA";
char * str2 = "Brandon Chong is the best TA";

The strings pointed to by str1 and str2 may actually reside in the same location in memory.

Char arrays, however, contain the literal value which has been copied from the code segment into either the stack or static memory. These following char arrays do not reside in the same place in memory.

char arr1[] = "Brandon Chong didn't write this";
char arr2[] = "Brandon Chong didn't write this";

Buffer overflow/ underflow

#define N (10)
int i = N, array[N];
for( ; i >= 0; i--) array[i] = i;

C does not check that pointers are valid. The above example writes into array[10] which is outside the array bounds. This can cause memory corruption because that memory location is probably being used for something else. In practice, this can be harder to spot because the overflow/underflow may occur in a library call e.g.

gets(array); // Let's hope the input is shorter than my array!

Returning pointers to automatic variables

int *f() {
    int result = 42;
    static int imok;
    return &imok; // OK - static variables are not on the stack
    return &result; // Not OK
}

Automatic variables are bound to stack memory only for the lifetime of the function. After the function returns it is an error to continue to use the memory.

Insufficient memory allocation

struct User {
   char name[100];
};
typedef struct User user_t;

user_t *user = (user_t *) malloc(sizeof(user));

In the above example, we needed to allocate enough bytes for the struct. Instead we allocated enough bytes to hold a pointer. Once we start using the user pointer we will corrupt memory. Correct code is shown below.

struct User {
   char name[100];
};
typedef struct User user_t;

user_t * user = (user_t *) malloc(sizeof(user_t));

Strings require strlen(s)+1 bytes

Every string must have a null byte after the last characters. To store the string "Hi" it takes 3 bytes: [H] [i] [\0].

  char *strdup(const char *s) {  /* return a copy of 'input' */
    char *copy;
    copy = malloc(sizeof(char*));     /* nope! this allocates space for a pointer, not a string */
    copy = malloc(strlen(input));     /* Almost...but what about the null terminator? */
    copy = malloc(strlen(input) + 1); /* That's right. */
    strcpy(copy, input);   /* strcpy will provide the null terminator */
    return copy;
}

Using uninitialized variables

int myfunction() {
  int x;
  int y = x + 2;
...

Automatic variables hold garbage (whatever bit pattern happened to be in memory). It is an error to assume that it will always be initialized to zero.

Assuming Uninitialized memory will be zeroed

void myfunct() {
   char array[10];
   char *p = malloc(10);

Automatic (temporary variables) are not automatically initialized to zero. Heap allocations using malloc are not automatically initialized to zero.

Double-free

  char *p = malloc(10);
  free(p);
//  .. later ...
  free(p); 

It is an error to free the same block of memory twice.

Dangling pointers

  char *p = malloc(10);
  strcpy(p, "Hello");
  free(p);
//  .. later ...
  strcpy(p,"World"); 

Pointers to freed memory should not be used. A defensive programming practice is to set pointers to null as soon as the memory is freed.

It is a good idea to turn free into the following snippet that automatically sets the freed variable to null right after:(vim - ultisnips)

snippet free "free(something)" b
free(${1});
$1 = NULL;
${2}
endsnippet

Logic and Program flow mistakes

Forgetting break

int flag = 1; // Will print all three lines.
switch(flag) {
  case 1: printf("I'm printed\n");
  case 2: printf("Me too\n");
  case 3: printf("Me three\n");
}

Case statements without a break will just continue onto the code of the next case statement. Correct code is show bellow. The break for the last statements is unnecessary because there are no more cases to be executed after the last one. However if more are added, it can cause some bugs.

int flag = 1; // Will print all three lines.
switch(flag) {
  case 1: 
    printf("I'm printed\n");
    break;
  case 2: 
    printf("Me too\n");
    break;
  case 3: 
    printf("Me three\n");
    break; //unnecessary
}

Equal vs. equality

int answer = 3; // Will print out the answer.
if (answer = 42) { printf("I've solved the answer! It's %d", answer);}

Undeclared or incorrectly prototyped functions

time_t start = time();

The system function 'time' actually takes a parameter (a pointer to some memory that can receive the time_t structure). The compiler did not catch this error because the programmer did not provide a valid function prototype by including time.h

Extra Semicolons

for(int i = 0; i < 5; i++) ; printf("I'm printed once");
while(x < 10); x++ ; // X is never incremented

However, the following code is perfectly OK.

for(int i = 0; i < 5; i++){
    printf("%d\n", i);;;;;;;;;;;;;
}

It is OK to have this kind of code, because the C language uses semicolons (;) to separate statements. If there is no statement in between semicolons, then there is nothing to do and the compiler moves on to the next statement

Other Gotchas

C Preprocessor macros and side-effects

#define min(a,b) ((a)<(b) ? (a) : (b))
int x = 4;
if(min(x++, 100)) printf("%d is six", x);

Macros are simple text substitution so the above example expands to x++ < 100 ? x++ : 100 (parenthesis omitted for clarity)

C Preprocessor macros and precedence

#define min(a,b) a<b ? a : b
int x = 99;
int r = 10 + min(99, 100); // r is 100!

Macros are simple text substitution so the above example expands to 10 + 99 < 100 ? 99 : 100

Assignments in Conditions

int a = 0;
if (a = 1) {
    printf("What is a?\n");
}

Notice the second line--a = 1 vs. a == 1. What happens here? The assignment operator in C returns the value on the right. So in this case, if (a = 1) evaluates to if (1).