It's supposed to be an infinite loop, but it stops after execv

Question

I'm developing a shell command in C, and I have a problem when using the command execv. I have a do while(1) in the main, so it should be an infinite loop, but if i introduce /bin/ls as the command, the prompt shows the files and folders and then it stops.

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <errno.h>
#include <sys/types.h>
#include <sys/wait.h>

#define MAX_LENGTH 256
#define DEFAULT_STRING_LENGTH  256
#define MAX_PARAMETERS  16


void  initParams(char  ***  params);
void  read_command(char  *** params);
void  freeParams(char  ***  params);
void type_prompt();
int comprobarSalir(char ***cadena);

int main(){

    int salir = 0;

    do{


    char ** params;



    type_prompt();

    fflush(stdout);
    initParams (& params);
    read_command (& params);
    salir = comprobarSalir(& params);

    if(comprobarSalir(&params)==0){

        if(execv(params[0],  params) ==  -1){
            printf("%s,%s",params[0],params[1]);
            printf("Error  al  ejecutar  el  comando ' %s':  %s\n", params [0],  strerror(errno));
            freeParams (& params);
        }

    }


    }while(1);


}


void  read_command(char  *** args) {
    char  input [256], *substr;
    int n = 0;
    fgets(input , sizeof(input), stdin);
    input[strlen(input) -1] = '\0';
    substr = strtok(input , " ");
    if (substr  != NULL)
        memcpy ((* args)[n], substr , strlen(substr));
    else
        (*args)[n] = NULL;
        n++;
    while ((* args)[n-1] != NULL) {
        substr = strtok(NULL , " ");
        if (substr  != NULL)
            memcpy ((* args)[n], substr , strlen(substr));
        else
            (*args)[n] = NULL;
            n++;
    }
}

void  freeParams(char  *** params){
    int i;
    char ** parameter;
    for (i=0; i<MAX_PARAMETERS; i++) {
        parameter = ((* params) +i);
        if (* parameter  != NULL) free(* parameter);
    }
    free(* params);
    *params = NULL;
}


void  initParams(char  *** params) {
    int i, j;
    char ** parameter;
    *params = (char **) malloc(sizeof(parameter) * MAX_PARAMETERS);
    for (i = 0; i<MAX_PARAMETERS; i++) {
        parameter = (* params) + i;
        *parameter = (char*) malloc(DEFAULT_STRING_LENGTH);
        for (j = 0; j<DEFAULT_STRING_LENGTH; j++)  *((* parameter)+j) = '\0';
    }
}

void type_prompt(){
    char cwd[MAX_LENGTH];
    getcwd(cwd, sizeof(cwd));
    printf("%s$ ",cwd);
}

int comprobarSalir(char ***cadena){

    int salir = 0;
    char* exit = "exit";

    if(***cadena==*exit){
        salir = 1231;
    }

    return salir;
}

Answer 1

execv replaces the process that invokes it with a different program.

Since the current process is now running a different program (ls), of course its loop will no longer continue.

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If you want a process to continue to run after calling execv, then you need to use fork() to create a child process; check whether you're the parent or the child, and call execv only in the child. At that point, if you want the parent to wait for the copy of ls to finish, you can use wait() in the parent to do so, and determine whether the child succeeded or failed.