将字符串转换为int数组

Question

I have a following problem. I want to convert a string like "10 15 30" to an int array with ints 10, 15, 30 . I searched in google a lot, but usually solutions included vectors (which I am not familiar with) or other more complex solutions. I found a code like this:

#include <cstdio>
void foo ( char *line ) {
    int num, i = 0, len;
    while ( sscanf( line, "%d%n", &num, &len) == 1 ) {
        printf( "Number %d is %d; ", i, num );
        line += len;    
        i++;            
    }
}
int main ( ) {
    char test[] = "12 45 6 23 100";
    foo( test );
    return 0;
}

It works and extracts numbers from string in a way I wanted, but I don't understand part with:

line += len;

Can someone explain how it works? Why are we adding len (which is int ) to the string?

Answer 1

Solution for C++:

#include <iostream>
#include <sstream>
#include <vector>

std::vector< int > foo ( char *c_str ) {
    std::istringstream line( c_str );
    std::vector< int > numbers;
    for ( int n; line >> n; )
        numbers.push_back( n );
    return numbers;
}

int main ( ) {
    char test[] = "12 45 6 23 100";
    std::vector< int > numbers = foo( test );
    for ( int n : numbers )
        std::cout << n << ' ';
    return 0;
}

Output:

12 45 6 23 100

Answer 2

%n specifies the number of characters accessed by sscanf in its one execution and saved it in variable len. so line + =len; is incrementing variable line , with number of characters accessed by sscanf

line + =len; is nothing but line =line+ len;

Answer 3

sscanf( line, "%d%n", &num, &len)

This line reads a number from line, and put the digit into num and number of characters read into len.

We need len to move pointer to unread portion of the string.

Does it make sense ?

Useful links: scanf sscanf

Answer 4

line += len; is equivalent to line = line + len;

line is a char pointer, thus the code increases the pointer by len , so that the pointer points to the next number, every time the loop is executed. Note than len is updated by the call to sscanf() .

Answer 5

The "string" is a pointer to a character buffer. You are performing pointer arithmetic to increment the sscanf to parse the next int characters from the buffer. So, in your example:

char test[] = "12 45 6 23 100";

say *line points to test, and has some pointer value, we don't care what the value is per-se, so let's say it is 1000. After the 1st call to sscanf, you read "12" in. That is two characters, so len should return 2. line then gets set to 1002, and is now "pointing" to the next set of characters in the buffer. sscanf reads that, and this continues until no more characters in the buffer.

Answer 6

Look at sscanf docs . It receive format and additional params.

In params you have "%d%n" which you explain using format table from docs .

d or u

Decimal integer

Any number of decimal digits (0-9), optionally preceded by a sign (+ or -). d is for a signed argument, and u for an unsigned.

n

Count

No input is consumed. The number of characters read so far from stdin is stored in the pointed location.

So you read integer from string and number of chars this integer consist from.

After reading part of string you move pointer into number of read chars.