如何在c中的同一行打印两个字符串

Question

I want to input a string with spaces and print that string with a another string on the same line.

int i = 4;
double d = 4.0;
char s[] = "Apple ";
int x;
double y;
char z[105];

scanf("%d",&x);
scanf("%lf",&y);
scanf("%[^105\n]s",z);

printf("%d\n",i+x);
printf("%0.1lf\n",d+y);
printf("%s %s",s,z);
return 0;

Answer 1

You scanf format specifier "%[^105]s" uses a character class [...] which is a stand-alone specifier in and of itself and does not requires 's' at the end. By placing 's' at the end you are forcing scanf to look for a literal 's' following an unlimited number of characters NOT including 1, 0, 5 .

It appears you intended to use the number to protect your arrays bounds -- which is a good thing, but the proper format in that case is "%104[^\\n]" which will read up to 104 characters that do not include a '\\n' (preserving space for the nul-character ).

For example:

    if (scanf("%104[^\n]",z) == 1)
        printf("%s %s\n",s,z);

( note: ALWAYS validate ALL user-input by at minimum checking the return)

Also note: by NOT reading the '\\n' above, it is left in your input buffer ( stdin ) unread, and if your next attempted input is "%c" or "%[...]" , you will take the '\\n' as part of your input as nether "%c" or `"%[...]" consume leading whitespace.

Putting it together in an example you could do:

#include <stdio.h>

int main (void) {
    char s[] = "Apple";
    char z[105];

    printf ("enter z: ");
    if (scanf("%104[^\n]",z) == 1)
        printf("%s %s\n",s,z);
    else
        fputs ("error: stream error or user canceled.\n", stderr);

    return 0;
}

( note: instead of scanf for reading lines, fgets() is recommended, then simply trim the '\\n' included in the filled buffer)

Example Use/Output

$ ./bin/oneline
enter z: is a fruit
Apple is a fruit

Use fgets() Instead

Instead of using scanf for line input, use a line-oriented input function like fgets() which will consume an entire line (including the line ending). The ensures your input buffer is left in a consistent state that does not depend on the previous format specifier user, eg

...
#include <string.h>
...
    printf ("enter z: ");
    if (fgets (z, sizeof z, stdin) != NULL) {
        z[strcspn (z, "\n")] = 0;               /* trim '\n' from end of z */
        printf("%s %s\n",s,z);
    }

Edit Per-Question in Comment

Your Problem With Your New Code Is scanf("%lf",&y); leaves the '\\n' in stdin unread , you then attempt to read scanf("%[^105\\n]",z); which reads nothing because you have excluded reading '\\n' in the inverted character class and you then read stdin as input where the first character is '\\n' . "%[^105\\n]" means : read an unlimited number of characters and only stop the read if a 1, 0, 5 or '\\n' character (or EOF ) is encountered.

Taking mixed input with scanf is full of Pitfalls for new C programmers because of what is left in stdin , and how leading whitespace is handled depends on the format-specifier used. This is why fgets() (or POSIX getline() ) are recommended for user input, and then parsing the needed information from the filled buffer with sscanf . With a line-oriented input function, the line is completely consumed on each input (given a sufficient buffer size -- don't skimp), eliminating the problems with scanf .

To make your current code work, you could do:

#include <stdio.h>

/* simple function to empty remainder of line in stdin */
void empty_stdin (void)
{
    int c = getchar();

    while (c != '\n' && c != EOF)
        c = getchar();
}

int main (void) {

    int i = 4, x; 
    double d = 4.0, y; 
    char s[] = "Apple ", z[105];

    scanf("%d",&x);
    scanf("%lf",&y);        /* leaves '\n' as next char in stdin */
    empty_stdin();          /* empty extraneous characters */
    scanf("%104[^\n]",z);   /* read up to 104 chars, \n, or EOF */

    printf("%d\n",i+x);
    printf("%0.1lf\n",d+y);
    printf("%s %s\n",s,z);

    return 0;
}

( validate each call to scanf -- that is left to you)

Let me know if you have further questions.

Answer 2

我在scanf中使用\\ n解决了这个问题（ “％lf \\ n ”，＆y）;