Question

到目前为止，我只处理第一个显示十六进制偏移量＆＃34; 0＆＃34;我试图找到一种程序打开文件并显示带有前导0x指示符的8位十六进制偏移量的方法。我试图让它像这样运行：

----------------------------------------------------------------------------- 
 offset      0  1  2  3   4  5  6  7   8  9  a  b   c  d  e  f  ascii
-----------------------------------------------------------------------------
0x00000000
0x00000010 
0x00000020

依旧......

任何提示/建议都会很棒。此外，我也在寻找一个＆＃34;清洁＆＃34;显示横幅的方式，而不是 printf 各个行等等。

所以，我的问题是：

如何让程序显示带有前导0x的十六进制偏移量。我尝试为该部分插入我的代码，但我无法确定如何，所以任何提示都会受到赞赏。
有没有更简洁的方式来显示标题？

非常感谢你。

#include <stdio.h>
#include <stdlib.h>
#include <sys/ioctl.h>


int main(int argc, char** argv)
{
    struct winsize terminal;
    FILE* fp;
    int a, c, column, row, lines;

    ioctl(0, TIOCGWINSZ, &terminal, argv[2]);

    fp = fopen(argv[1], "r");                          //set open for argv[1] and set it to read mode
    column = terminal.ws_col;
    row = terminal.ws_row;
    lines = atoi(argv[2]);

    // printf("Rows are: %d\n", row);
    // printf("Columns are: %d\n", column);

    if (argc < 3)                                      //Argument checking, at least three to run
    {
        printf("You need atleast 3 arguments to run!\n");
        exit(1);
    }

    if (fp == NULL)
    {
        fprintf(stderr, "Error opening '%s'.\n", argv[1]);             //if not file is specified then close program and give an error
        exit(1);
    }
    c = fgetc(fp);                                //set c to fgetc

    if (column < 80 || row < 20)                               //if temrinal window is less than 80x20, display an error and exit
    {
        printf("Terminal window must be at least 80x20!\n");              //display error and close program if column criteria isn't met
        exit(1);
    }

    /*
    while(!feof(fp))                                   //if file is entered, display a test message (this is where later on I will show hex value)

    {
        printf("Good Job! You picked a file to manipulate!!\n");
        exit(1);
     }
     */
    if (lines == 0)                                       //if 0 is entered for argv[2], display all the lines, 512 byte file
    {
        printf("------------------------------------------------------------------------------\n"); //**this is where I try to display character/hex banner**
        printf("offset      0  1  2  3   4  5  6  7   8  9  a  b   c  d  e  f  ascii\n");
        printf("-------------------------------------------------------------------------------\n");
        for (c == 0x200) ; //**This is where I try to get the 8 digit 0x hex offset**
        {
            fprintf(stdout, "%04x", c);
        }
    }
    else if (lines == 20)
    {
        printf("This will print out 20 lines\n");
    }
    else if (lines == 30)
    {
        printf("This will print out 30 lines\n");
    }
    else
    {
        printf("Please Enter a Valid Number!\n");
    }

    fclose(fp);                                        //close fp

    return 0;
}

Answer 1

尝试

printf("0x%08x\n", 0xDEADBEEF);

显示8位十六进制偏移量0x？

1 个答案: