尝试在程序中输入少于5个参数时,我一直遇到“分段错误”。我知道Java,但是对C还是陌生的,只是不确定发生了什么。我只是尝试将用户输入的kg(作为参数)转换为Prius的重量(十进制)。
#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>
int main(int argc, char *argv[]) //Argc is the argument count, argv is the argument vector
{
//Initialize the toyota/user variables
int toyota = 1325; //Toyota weight in kg
int paramOne = atoi(argv[1]);
int paramTwo = atoi(argv[2]);
int paramThree = atoi(argv[3]);
int paramFour = atoi(argv[4]);
//This if will check to see if there are too many parameters
if (argc >= 5)
{
printf("Error: Too many parameters.\n");
printf("Usage: ./HW1 arg1 [arg2 ... arg4].\n");
}
//This if will check to see if there are too few parameters
if (argc < 2)
{
printf("Error: Too few parameters.\n");
printf("Usage: ./HW1 arg1 [arg2 ... arg4.\n");
}
//If there are a correct amount of parameters, this will print the TP count
if ((argc >= 1) && (argc <= 4))
{
printf("%d number of parameters.\n", argc);
if(argc >= 1)
{
printf("%d kg = %d TP.\n", paramOne, paramOne/toyota); //Parameter divided by TP
}
if(argc >= 2)
{
printf("%d kg = %d TP.\n", paramTwo, paramTwo/toyota);
}
if(argc >= 3)
{
printf("%d kg = %d TP.\n", paramThree, paramThree/toyota);
}
if(argc >= 4)
{
printf("%d kg = %d TP.\n", paramFour, paramFour/toyota);
}
}
}
最佳答案
int paramOne = atoi(argv[1]);
int paramTwo = atoi(argv[2]);
int paramThree = atoi(argv[3]);
int paramFour = atoi(argv[4]);
如果是
argc <= 4
,则这些调用会调用 undefined 的行为。请先测试argc
,如果参数不足,请不要调用atoi
。