我的程序在不同的机器上提供不同的输出..!

Cap*_*owl 6 c c++ string concatenation

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


int main()
{
    char left[4];
    for(int i=0; i<4; i++)
    {
        left[i]='0';
    }
    char str[10];
    gets(str);
    strcat(left,str);
    puts(left);
    return 0;
}
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对于任何输入,它应该连接0000与该字符串,但在一台电脑上它显示"0000"和输入字符串之间的菱形标志......!

Som*_*ude 6

您可以将九个(或更多,gets没有边界检查)字符串附加到三个字符的字符串(其中包含四个字符且没有字符串终止符).根本没有字符串终止.因此,当您使用puts它进行打印时,它将继续打印,直到找到字符串终止字符,这可能是内存中的任何位置.简而言之,这是缓冲区溢出的学校书籍示例,缓冲区溢出通常会导致未定义的行为,这正是您所看到的.

在C和C++中,必须终止所有C风格的字符串.它们由特殊字符终止:( '\0'或纯ASCII零).您还需要在strcat通话中为目标字符串提供足够的空间.


适当的工作计划:

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

int main(void)
{
    /* Size is 4 + 10 + 1, the last +1 for the string terminator */
    char left[15] = "0000";
    /* The initialization above sets the four first characters to '0'
     * and properly terminates it by adding the (invisible) '\0' terminator
     * which is included in the literal string.
     */

    /* Space for ten characters, plus terminator */
    char str[11];

    /* Read string from user, with bounds-checking.
     * Also check that something was truly read, as `fgets` returns
     * `NULL` on error or other failure to read.
     */
    if (fgets(str, sizeof(str), stdin) == NULL)
    {
        /* There might be an error */
        if (ferror(stdin))
            printf("Error reading input: %s\n", strerror(errno));
        return 1;
    }

    /* Unfortunately `fgets` may leave the newline in the input string
     * so we have to remove it.
     * This is done by changing the newline to the string terminator.
     *
     * First check that the newline really is there though. This is done
     * by first making sure there is something in the string (using `strlen`)
     * and then to check if the last character is a newline. The use of `-1`
     * is because strings like arrays starts their indexing at zero.
     */
    if (strlen(str) > 0 && str[strlen(str) - 1] == '\n')
        str[strlen(str) - 1] = '\0';

    /* Here we know that `left` is currently four characters, and that `str` 
     * is at most ten characters (not including zero terminaton). Since the
     * total length allocated for `left` is 15, we know that there is enough
     * space in `left` to have `str` added to it.
     */
    strcat(left, str);

    /* Print the string */
    printf("%s\n", left);

    return 0;
}
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  • 我称之为"缓冲区溢出*的学校书籍示例".不太通用的术语. (3认同)