In their book The Practice of Programming (which is well worth reading), Kernighan and Pike discuss this problem, and they solve it by using snprintf() to create the string with the correct buffer size for passing to the scanf() family of functions. In effect: int scanner(const char *data, char *buffer, size_t buflen) { char format[32]; … Read more
The most common ways of reading input are: using fgets with a fixed size, which is what is usually suggested, and using fgetc, which may be useful if you’re only reading a single char. To convert the input, there are a variety of functions that you can use: strtoll, to convert a string into an … Read more
scanf consumes only the input that matches the format string, returning the number of characters consumed. Any character that doesn’t match the format string causes it to stop scanning and leaves the invalid character still in the buffer. As others said, you still need to flush the invalid character out of the buffer before you … Read more
People (and especially beginners) should never use scanf(“%s”) or gets() or any other functions that do not have buffer overflow protection, unless you know for certain that the input will always be of a specific format (and perhaps not even then). Remember than scanf stands for “scan formatted” and there’s precious little less formatted than … Read more
A whitespace character in a scanf format causes it to explicitly read and ignore as many whitespace characters as it can. So with scanf(“%d “, …, after reading a number, it will continue to read characters, discarding all whitespace until it sees a non-whitespace character on the input. That non-whitespace character will be left as … Read more
scanf() reads exactly what you asked it to, leaving the following \n from the end of that line in the buffer where fgets() will read it. Either do something to consume the newline, or (my preferred solution) fgets() and then sscanf() from that string.
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don’t skip whitespace. Use ” %c” with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() … Read more
Three problems here. First scanf(ifp,”%lf %lf\n”,&theta,&Cla); You’re calling scanf when you look like you want to use fscanf: fscanf(ifp,”%lf %lf\n”,&theta,&Cla); Second is the file you’re opening: ifp=fopen(argv[0],”r”); argv[0] is the name of the running executable, which is probably not what you want. If you want the first argument passed in, use argv[1] ifp=fopen(argv[1],”r”); Last is … Read more
The code contains a typo. In this statement scand(“%f”,&Centimeters); there shall be scanf instead of scand. scanf(“%f”,&Centimeters);
@n.Doe, What you might be experiencing is that the variables aren’t being initialized and thus see some random values when the sscanf() doesn’t get to find the content for those missing ones. Try initializing them as following. int int1 = 0, int2 = 0, int3 = 0; Also, the sscanf() needs the variables addresses (hence … Read more