This document lists the common C programming errors that the author sees time and time again. Solutions to the errors are also presented.
2. Beginner Errors
These are errors that beginning C students often make. However, the professionals still sometimes make them too!
- Forgetting to put a break in a switch statement
Remember that C does not break out of a switch statement if a case is encountered. For example:
int x = 2;
switch(x) {
case 2:
printf("Two\n");
case 3:
printf("Three\n");
}
prints out:
Two
Three
Put a break to break out of the switch:
int x = 2;
switch(x) {
case 2:
printf("Two\n");
break;
case 3:
printf("Three\n");
break; /* not necessary, but good if additional cases are added later */
}
- Using = instead of ==
C's = operator is used exclusively for assignment and returns the value assigned. The == operator is used exclusively for comparison and returns an integer value (0 for false, not 0 for true). Because of these return values, the C compiler often does not flag an error when = is used when one really wanted an ==. For example:
int x = 5;
if ( x = 6 )
printf("x equals 6\n");
This code prints out x equals 6! Why? The assignment inside the if sets x to 6 and returns the value 6 to the if. Since 6 is not 0, this is interpreted as true.
Avoiding this error -
One way to have the compiler find this type of error is to put any constants (or any r-value expressions) on the left side. Then if an = is used, it will be an error:
if ( 6 = x)
- scanf() errors
There are two types of common scanf() errors:
2.3.1 Forgetting to put an ampersand (&) on arguments
scanf() must have the address of the variable to store input into. This means that often the ampersand address operator is required to compute the addresses. Here's an example:
int x;
char * st = malloc(31);
scanf("%d", &x); /* & required to pass address to scanf() */
scanf("%30s", st); /* NO & here, st itself points to variable! */
As the last line above shows, sometimes no ampersand is correct!
2.3.2 Using the wrong format for operand
C compilers do not check that the correct format is used for arguments of a scanf() call. The most common errors are using the %f format for doubles (which must use the %lf format) and mixing up %c and %s for characters and strings.
- Size of arrays
Arrays in C always start at index 0. This means that an array of 10 integers defined as:
int a[10];
has valid indices from 0 to 9 not 10! It is very common for students go one too far in an array. This can lead to unpredictable behavior of the program.
2.5 Integer division
Unlike Pascal, C uses the / operator for both real and integer division. It is important to understand how C determines which it will do. If both operands are of an integal type, integer division is used, else real division is used. For example:
double half = 1/2;
This code sets half to 0 not 0.5! Why? Because 1 and 2 are integer constants. To fix this, change at least one of them to a real constant.
double half = 1.0/2;
If both operands are integer variables and real division is desired, cast one of the variables to double (or float).
int x = 5, y = 2;
double d = ((double) x)/y;
2.6 Loop errors
In C, a loop repeats the very next statement after the loop statement. The code:
int x = 5;
while( x > 0 );
x--;
is an infinite loop. Why? The semicolon after the while defines the statement to repeat as the null statement (which does nothing). Remove the semicolon and the loop works as expected.
Another common loop error is to iterate one too many times or one too few. Check loop conditions carefully!
2.7 Not using prototypes
Prototypes tell the compiler important features of a function: the return type and the parameters of the function. If no prototype is given, the compiler assumes that the function returns an int and can take any number of parameters of any type.
One important reason to use prototypes is to let the compiler check for errors in the argument lists of function calls. However, a prototype must be used if the function does not return an int. For example, the sqrt() function returns a double, not an int. The following code:
double x = sqrt(2);
will not work correctly if a prototype:
double sqrt(double);
does not appear above it. Why? Without a prototype, the C compiler assumes that sqrt() returns an int. Since the returned value is stored in a double variable, the compiler inserts code to convert the value to a double. This conversion is not needed and will result in the wrong value.
The solution to this problem is to include the correct C header file that contains the sqrt() prototype, math.h. For functions you write, you must either place the prototype at the top of the source file or create a header file and include it.
2.8 Not initializing pointers
Anytime you use a pointer, you should be able to answer the question: What variable does this point to? If you can not answer this question, it is likely it doesn't point to any variable. This type of error will often result in a Segmentation fault/coredump error on UNIX/Linux or a general protection fault under Windows. (Under good old DOS (ugh!), anything could happen!)
Here's an example of this type of error.
#include
int main()
{
char * st; /* defines a pointer to a char or char array */
strcpy(st, "abc"); /* what char array does st point to?? */
return 0;
}
How to do this correctly? Either use an array or dynamically allocate an array.
#include
int main()
{
char st[20]; /* defines an char array */
strcpy(st, "abc"); /* st points to char array */
return 0;
}
or
#include
#include
int main()
{
char *st = malloc(20); /* st points to allocated array*/
strcpy(st, "abc"); /* st points to char array */
free(st); /* don't forget to deallocate when done! */
return 0;
}
Actually, the first solution is much preferred for what this code does. Why? Dynamical allocation should only be used when it is required. It is slower and more error prone than just defining a normal array.
3. String Errors
3.1 Confusing character and string constants
C considers character and string constants as very different things. Character constants are enclosed in single quotes and string constants are enclosed in double quotes. String constants act as a pointer to the actually string. Consider the following code:
char ch = 'A'; /* correct */
char ch = "A"; /* error */
The second line assigns the character variable ch to the address of a string constant. This should generate a compiler error. The same should happen if a string pointer is assigned to a character constant:
const char * st = "A"; /* correct */
const char * st = 'A'; /* error */
3.2 Comparing strings with ==
Never use the == operator to compare the value of strings! Strings are char arrays. The name of a char array acts like a pointer to the string (just like other types of arrays in C). So what? Consider the following code:
char st1[] = "abc";
char st2[] = "abc";
if ( st1 == st2 )
printf("Yes");
else
printf("No");
This code prints out No. Why? Because the == operator is comparing the pointer values of st1 and st2, not the data pointed to by them. The correct way to compare string values is to use the strcmp() library function. (Be sure to include string.h) If the if statement above is replaced with the following:
if ( strcmp(st1,st2) == 0 )
printf("Yes");
else
printf("No");
the code will print out Yes. For similar reasons, don't use the other relational operators (<,>, etc.) with strings either. Use strcmp() here too.
3.3 Not null terminating strings
C assumes that a string is a character array with a terminating null character. This null character has ASCII value 0 and can be represented as just 0 or '\0'. This value is used to mark the end of meaningful data in the string. If this value is missing, many C string functions will keep processing data past the end of the meaningful data and often past the end of the character array itself until it happens to find a zero byte in memory!
Most C library string functions that create strings will always properly null terminate them. Some do not (e.g., strncpy() ). Be sure to read their descriptions carefully.
3.4 Not leaving room for the null terminator
A C string must have a null terminator at the end of the meaningful data in the string. A common mistake is to not allocate room for this extra character. For example, the string defined below
char str[30];
only has room for only 29 (not 30) actually data characters, since a null must appear after the last data character.
This can also be a problem with dynamic allocation. Below is the correct way to allocate a string to the exact size needed to hold a copy of another.
char * copy_str = malloc( strlen(orig_str) + 1);
strcpy(copy_str, orig_str);
The common mistake is to forget to add one to the return value of strlen(). The strlen() function returns a count of the data characters which does not include the null terminator.
This type of error can be very hard to detect. It might not cause any problems or only problems in extreme cases. In the case of dynamic allocation, it might corrupt the heap (the area of the program's memory used for dynamic allocation) and cause the next heap operation (malloc(), free(), etc.) to fail.
4. Input/Output Errors
4.1 Using fgetc(), etc. incorrectly
The fgetc(), getc() and getchar() functions all return back an integer value. For example, the prototype of fgetc() is:
int fgetc( FILE * );
Sometimes this integer value is really a simple character, but there is one very important case where the return value is not a character!
What is this value? EOF A common misconception of students is that files have a special EOF character at the end. There is no special character stored at the end of a file. EOF is an integer error code returned by a function. Here is the wrong way to use fgetc():
int count_line_size( FILE * fp )
{
char ch;
int cnt = 0;
while( (ch = fgetc(fp)) != EOF && ch != '\n')
cnt++;
return cnt;
}
What is wrong with this? The problem occurs in the condition of the while loop. To illustrate, here is the loop rewritten to show what C will do behind the scenes.
while( (int) ( ch = (char) fgetc(fp) ) != EOF && ch != '\n')
cnt++;
The return value of fgetc(fp) is cast to char to store the result into ch. Then the value of ch must be cast back to an int to compare it with EOF. So what? Casting an int value to a char and then back to an int may not give back the original int value. This means in the example above that if fgetc() returns back the EOF value, the casting may change the value so that the comparison later with EOF would be false.
What is the solution? Make the ch variable an int as below:
int count_line_size( FILE * fp )
{
int ch;
int cnt = 0;
while( (ch = fgetc(fp)) != EOF && ch != '\n')
cnt++;
return cnt;
}
Now the only hidden cast is in the second comparison.
while( (ch = fgetc(fp)) != EOF && ch != ((int) '\n') )
cnt++;
This cast has no harmful effects at all! So, the moral of all this is: always use an int variable to store the result of the fgetc(), getc() and getchar().
4.2 Using feof() incorrectly
There is a wide spread misunderstanding of how C's feof() function works. Many programmers use it like Pascal's eof() function. However, C's function works differently!
What's the difference? Pascal's function returns true if the next read will fail because of end of file. C's function returns true if the last function failed. Here's an example of a misuse of feof():
#include
int main()
{
FILE * fp = fopen("test.txt", "r");
char line[100];
while( ! feof(fp) ) {
fgets(line, sizeof(line), fp);
fputs(line, stdout);
}
fclose(fp);
return 0;
}
This program will print out the last line of the input file twice. Why? After the last line is read in and printed out, feof() will still return 0 (false) and the loop will continue. The next fgets() fails and so the line variable holding the contents of the last line is not changed and is printed out again. After this, feof() will return true (since fgets() failed) and the loop ends.
How should this fixed? One way is the following:
#include
int main()
{
FILE * fp = fopen("test.txt", "r");
char line[100];
while( 1 ) {
fgets(line, sizeof(line), fp);
if ( feof(fp) ) /* check for EOF right after fgets() */
break;
fputs(line, stdout);
}
fclose(fp);
return 0;
}
However, this is not the best way. There is really no reason to use feof() at all. C input functions return values that can be used to check for EOF. For example, fgets returns the NULL pointer on EOF. Here's a better version of the program:
#include
int main()
{
FILE * fp = fopen("test.txt", "r");
char line[100];
while( fgets(line, sizeof(line), fp) != NULL )
fputs(line, stdout);
fclose(fp);
return 0;
}
The author has yet to see any student use the feof() function correctly!
Incidently, this discussion also applies to C++ and Java. The eof() method of an istream works just like C's feof().
4.3 Leaving characters in the input buffer
C input (and output) functions buffer data. Buffering stores data in memory and only reads (or writes) the data from (or to) I/O devices when needed. Reading and writing data in big chunks is much more efficient than a byte (or character) at a time. Often the buffering has no effect on programming.
One place where buffering is visible is input using scanf(). The keyboard is usually line buffered. This means that each line input is stored in a buffer. Problems can arise when a program does not process all the data in a line, before it wants to process the next line of input. For example, consider the following code:
int x;
char st[31];
printf("Enter an integer: ");
scanf("%d", &x);
printf("Enter a line of text: ");
fgets(st, 31, stdin);
The fgets() will not read the line of text that is typed in. Instead, it will probably just read an empty line. In fact, the program will not even wait for an input for the fgets() call. Why? The scanf() call reads the characters needed that represent the integer number read in, but it leaves the '\n' in the input buffer. The fgets() then starts reading data from the input buffer. It finds a '\n' and stops without needing any additional keyboard input.
What's the solution? One simple method is to read and dump all the characters from the input buffer until a '\n' after the scanf() call. Since this is something that might be used in lots of places, it makes sense to make this a function. Here is a function that does just this:
/* function dump_line
* This function reads and dumps any remaining characters on the current input
* line of a file.
* Parameter:
* fp - pointer to a FILE to read characters from
* Precondition:
* fp points to a open file
* Postcondition:
* the file referenced by fp is positioned at the end of the next line
* or the end of the file.
*/
void dump_line( FILE * fp )
{
int ch;
while( (ch = fgetc(fp)) != EOF && ch != '\n' )
/* null body */;
}
Here is the code above fixed by using the above function:
int x;
char st[31];
printf("Enter an integer: ");
scanf("%d", &x);
dump_line(stdin);
printf("Enter a line of text: ");
fgets(st, 31, stdin);
One incorrect solution is to use the following:
fflush(stdin);
This will compile but its behavior is undefined by the ANSI C standard. The fflush() function is only meant to be used on streams open for output, not input. This method does seem to work with some C compilers, but is completely unportable! Thus, it should not be used.
4.4 Using the gets() function
Do not use this function! It does not know how many characters can be safely stored in the string passed to it. Thus, if too many are read, memory will be corrupted. Many security bugs that have been exploited on the Internet use this fact! Use the fgets() function instead (and read from stdin). But remember that unlike gets(), fgets() does not discard a terminating \n from the input.
The scanf() functions can also be used dangerously. The %s format can overwrite the destination string. However, it can be used safely by specifying a width. For example, the format %20s will not read more than 20 characters.
5. Acknowlegements
The author would like to thank Stefan Ledent for suggesting the section on "Not leaving room for the null terminator"
