execvp（）无法在我的shell中工作-execvp()notworkinginmyshell

I am trying to make a tiny shell. My problem is that when I call execvp() - I get errors. For example, when I type in ls -l it returns ls: invalid option -- '

我想做一个小壳。我的问题是，当我调用execvp（）时 - 我得到错误。例如，当我输入ls -l时，它返回ls：invalid选项 - '

Can someone, please, help me understand why I am getting this error? For my code, the function command split gets the user input, and splits them up into separate commands. Separate commands are seperated by ; character.

请有人帮助我理解为什么我会收到此错误？对于我的代码，函数命令split获取用户输入，并将它们拆分为单独的命令。单独的命令分开;字符。

#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define MAX_CHARACTERS 512
#define HISTORY_SIZE 10

int commandSplit(char *c, char *a[], int t[]) {

int count = 0;
int total = 0;
char *temp[MAX_CHARACTERS];
char *readCommands = strtok(c, ";");
while(readCommands != NULL) {
    printf("Reading full command: %s\n", readCommands);
    temp[count] = readCommands;
    count++;
    readCommands = strtok(NULL, ";");
}
printf("Done reading full commands\n");
for(int i = 0; i   ");
fgets(commands, MAX_CHARACTERS, stdin);

if(strlen(commands) == 0) continue;

numOfCommands = commandSplit(commands, args, tracker);
printf("There are %i commands!\n", numOfCommands);

if(strcmp(args[0], "exit") == 0) {
    printf("Exiting\n");
    exitProgram = 1;
    continue;
}

int l = 0;
for(int i = 0; i 

int len;
len = strlen(commands);
if (len > 0 && commands[len-1] == '\n') {
    commands[len-1] = '\0';
}