Reading a string with spaces with sscanf

Question

For a project I'm trying to read an int and a string from a string. The only problem is sscanf() appears to break reading an %s when it sees a space. Is there anyway to get around this limitation? Here's an example of what I'm trying to do:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char** argv) {
    int age;
    char* buffer;
    buffer = malloc(200 * sizeof(char));
    sscanf("19 cool kid", "%d %s", &age, buffer);

    printf("%s is %d years old\n", buffer, age);
    return 0;
}

What it prints is: cool is 19 years old where I need cool kid is 19 years old . Does anyone know how to fix this?

Answer 1

以下行将开始读取一个数字（ %d ），然后是与制表符或换行符（ %[^\\t\\n] ）不同的任何内容。

sscanf("19 cool kid", "%d %[^\t\n]", &age, buffer);

Answer 2

You want the %c conversion specifier, which just reads a sequence of characters without special handling for whitespace.

Note that you need to fill the buffer with zeroes first, because the %c specifier doesn't write a nul-terminator. You also need to specify the number of characters to read (otherwise it defaults to only 1):

memset(buffer, 0, 200);
sscanf("19 cool kid", "%d %199c", &age, buffer);

Answer 3

If you want to scan to the end of the string (stripping out a newline if there), just use:

char *x = "19 cool kid";
sscanf (x, "%d %[^\n]", &age, buffer);

That's because %s only matches non-whitespace characters and will stop on the first whitespace it finds. The %[^\\n] format specifier will match every character that's not (because of ^ ) in the selection given (which is a newline). In other words, it will match any other character.

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Keep in mind that you should have allocated enough space in your buffer to take the string since you cannot be sure how much will be read (a good reason to stay away from scanf/fscanf unless you use specific field widths).

You could do that with:

char *x = "19 cool kid";
char *buffer = malloc (strlen (x) + 1);
sscanf (x, "%d %[^\n]", &age, buffer);

(you don't need * sizeof(char) since that's always 1 by definition).

Answer 4

Since you want the trailing string from the input, you can use %n (number of characters consumed thus far) to get the position at which the trailing string starts. This avoids memory copies and buffer sizing issues, but comes at the cost that you may need to do them explicitly if you wanted a copy.

const char *input = "19  cool kid";
int age;
int nameStart = 0;
sscanf(input, "%d %n", &age, &nameStart);
printf("%s is %d years old\n", input + nameStart, age);

outputs:

cool kid is 19 years old

Answer 5

I guess this is what you want, it does exactly what you specified.

#include<stdio.h>
#include<stdlib.h>

int main(int argc, char** argv) {
    int age;
    char* buffer;
    buffer = malloc(200 * sizeof(char));
    sscanf("19 cool kid", "%d cool %s", &age, buffer);
    printf("cool %s is %d years old\n", buffer, age);
    return 0;
}

The format expects: first a number (and puts it at where &age points to), then whitespace (zero or more), then the literal string "cool", then whitespace (zero or more) again, and then finally a string (and put that at whatever buffer points to). You forgot the "cool" part in your format string, so the format then just assumes that is the string you were wanting to assign to buffer. But you don't want to assign that string, only skip it.

Alternative, you could also have a format string like: "%d %s %s", but then you must assign another buffer for it (with a different name), and print it as: "%s %s is %d years old\\n".