Question

在程序下方获取用户输入，直到遇到一个空格。（程序与此一起工作）但我希望用户在仅按“Enter”键时停止它。所以我改变了

 (buffer[0] != '\0')

到

 (buffer[0] != '\n')

但这不起作用。我怎样才能做到这一点？（我知道“输入”意味着“\ n”）

#include <stdio.h>

int main(void){

int count = 0;
char buffer[80];

char *myDynamicArray[25];
int i = 0;
int k = 0;

puts("Enter integers, press space to end");

while ((i < 25) && (gets(buffer) != 0) && (buffer[0] != '\0'))
{
    if ((myDynamicArray[i] = (char *)malloc(strlen(buffer) + 1)) == NULL)
        return -1;
    strcpy(myDynamicArray[i++], buffer);

}

count = i;

for (k = 0; k < count; k++)
{
    printf("%s\n", myDynamicArray[k]);
}

getchar();

}

Answer 1

当用户点击＆＃34;输入＆＃34>时，您当前的程序会停止它。仅限密钥，如果您输入单个空格，它不会停止。

gets删除换行符并且不将其存储在缓冲区中，因此缓冲区[0]将等于＆＃39; \ 0＆＃39;没有输入，如果输入空格，则为。

一些注意事项：

启用编译器警告。警告为您提供潜在错误的提示。
您遗失了2个#include个文件：stdlib.h和string.h。
gets由于其不安全的性质visavi缓冲区溢出而被弃用。请改用fgets！

Answer 2

好的，我能想到的最基本的＃em>＆＃34;传统＆＃34; 基本方法就是这样做：

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

int main ( void )
{
    int c, i, j=0;
    char buffer[80], *dynArray[5];
    for (i=0;i<5;++i)
    {
        j = 0;
        while((c = getchar()) != EOF && c != '\n' && j < 79)
            buffer[j++] = c;
        if (j == 0)
        {//an empty string was encountered!
            for (--i;i>-1;--i)
            {//use --i, because current i is not allocated yet!
                printf("Freeing memory containing \"%s\"\n", dynArray[i]);
                free(dynArray[i]);
            }
            return EXIT_SUCCESS;//exit program here, then
        }
        buffer[j] = '\0';//add NULL-char
        dynArray[i] = calloc(j, sizeof *dynArray[i]);//calloc to initialize memory
        if (dynArray[i] == NULL)
            exit(EXIT_FAILURE);
        strcpy(dynArray[i], buffer);
    }
    //allow user to see a specific input string again:
    puts("Choose a string to see again [1-5], or enter q to proceed");
    while ((c = getchar()) != EOF && c != 'q')
    {
        c -= (1 + '0');//simple atoi trick
        if (c < 5 && c > -1)//input char is 1, 2, 3, 4 or 5
            printf("string %d: %s\n", c+1, dynArray[c]);
    }
    //q was fetched from stdin, clear buffer...
    while((c = getchar()) != EOF && c != '\n');
    for (i=0;i<5;++i)
    {
        printf("free-ing memory containing: %s\n", dynArray[i]);
        free(dynArray[i]);
    }
    getchar();//press any key to continue...
    return 0;
}

你可以随心所欲地玩这个（将其中的一些部分分成函数等）...
这是一个更有用的例子：

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

int get_key( const int max )
{
    int c, r = 0;
    while ((c = getchar()) != EOF && c != 'q' && c != '\n')
    {
        c -= '0';
        if (c > -1 && c < 10)
        {
            if (r > 0)//suppose input was 22, r will be 2 on second iteration of loop
                r *=10;//multiply by 10, to get 20
            r += c;//add 2 == 22
        }//optionally add else, clean buffer and procede if r > 0
    }
    if (r > max || r == 0)//perhaps move to loop body
        r = -1;//out-of-bounds request = -2
    if (c == 'q')
    {
        r = 0;//stop = -1
        //clean buffer
        while (c != EOF && c != '\n')
            c = getchar();
    }
    return r-1;//subtract 1 to get array index (zero indexed)
}

int main ( void )
{
    int c, k, i;
    char buffer[80], *myDynamicArray[25];
    puts("Enter string");
    for (k=0;k<25;++k)
    {
        i=0;
        while((c = getchar()) != EOF && c != '\n' && i < 79)
            buffer[i++] = c;
        if (i == 0)
            break;
        if (i == 79)
            while (c != EOF && c != '\n')
                c = getchar();//clean buffer
        buffer[i] = '\0';
        myDynamicArray[k] = calloc(i, sizeof *myDynamicArray[k]);
        if (myDynamicArray[k] == NULL)
            exit(EXIT_FAILURE);
        strcpy(myDynamicArray[k], buffer);
    }
    printf("Choose which string you'd like to see again [1-%d], press q to quit\n", k);
    while ((c = get_key(k)) != -1)
        if (c != -2)
            printf("String %d: %s\n", c+1, myDynamicArray[c]);
        else
            printf("input must be in the 1 - %d range\n", k);
    for (i=0;i<k;++i)
    {
        printf("%s\n", myDynamicArray[i]);
        free(myDynamicArray[i]);
    }
    printf("Press any key to continue...");
    getchar();
    return 0;
}

当用户按下＆＃34;输入＆＃34;如何结束此程序而不是一个空格字符？

2 个答案: