The accepted answer is incorrect. It will cause memory leaks.
Internally, yy_scan_string calls yy_scan_bytes which, in turn, calls yy_scan_buffer.
yy_scan_bytes allocates memory for a COPY of the input buffer.
yy_scan_buffer works directly upon the supplied buffer.
With all three forms, you MUST call yy_delete_buffer to free the flex buffer-state information (YY_BUFFER_STATE).
However, with yy_scan_buffer, you avoid the internal allocation/copy/free of the internal buffer.
The prototype for yy_scan_buffer does NOT take a const char* and you MUST NOT expect the contents to remain unchanged.
If you allocated memory to hold your string, you are responsible for freeing it AFTER you call yy_delete_buffer.
Also, don't forget to have yywrap return 1 (non-zero) when you're parsing JUST this string.
Below is a COMPLETE example.
%%
<<EOF>> return 0;
. return 1;
%%
int yywrap()
{
return (1);
}
int main(int argc, const char* const argv[])
{
FILE* fileHandle = fopen(argv[1], "rb");
if (fileHandle == NULL) {
perror("fopen");
return (EXIT_FAILURE);
}
fseek(fileHandle, 0, SEEK_END);
long fileSize = ftell(fileHandle);
fseek(fileHandle, 0, SEEK_SET);
char *string = malloc(fileSize + 2);
fread(string, fileSize, sizeof(char), fileHandle);
fclose(fileHandle);
string[fileSize] = '\0';
string[fileSize + 1] = '\0';
YY_BUFFER_STATE buffer = yy_scan_buffer(string, fileSize + 2);
int token;
do {
token = yylex();
} while (token != 0);
yy_delete_buffer(buffer);
free(string);
return (EXIT_SUCCESS);
}