C 通过链表循环-无止境循环

自学编程，我决定做一个简单的瓦特计算器。每个州有不同的（平均）电价/KWh。我的问题从代码中的第34行开始。curr->next从不为空

我做错了什么

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


float calculate(float w, float cents, float h);

struct states 
{
    char *name;
    char *price;
    struct states *next;
};

int main()
{
    float watts,hours,cents;
    int a = 0;
    struct states *curr, dict[52];
    char st_ab[52][3] = {"ID","ND","WA","AR","UT","LA","WY","SD","NE","KY","WV","OK","OR","MT","MS","TN","IN","NC","MS","IA","KS","VA","NM","MN","SC","AZ","GA","AL","CO","OH","TX","IL","FL","NV","MI","WI","PA","MD","DC","DE","MA","RI","CA","ME","VT","NJ","NH","AK","CT","NY","HI"};
    char st_pr[52][5] = {"8.0","8.1","8.2","8.7","8.8","8.9","8.9","9.0","9.0","9.0","9.2","9.2","9.4","9.6","9.7","9.8","10.0","10.2","10.3","10.5","10.5","10.5","10.7","10.9","11.0","11.1","11.1","11.1","11.2","11.2","11.3","11.7","11.7","11.7","13.0","13.0","13.2","13.7","13.7","13.7","14.8","14.9","15.2","15.5","16.1","16.3","16.5","17.5","18.1","18.1","33.2"};
    char state[2];

    for(a=0;a<=52;a++)
    {
        memset(&dict[a],0,sizeof(struct states)); /* zero out structure */
        if(a==52)
        {
            break;  /* if this is the last struct, leave it NULL so we can loop through linked list */
        }
        dict[a].next=&dict[a+1];        
    }

    for(curr = dict;curr ->next != NULL;curr = curr->next)  /* fill our list with state abbr. and cost/KWh */
    {
        curr->name = st_ab[a];
        curr->price = st_pr[a];
        a++;
    }

    puts("Dictionary loaded!");
    printf(" Two letter state code: ");                 /* get info from user */
    scanf("%s",state);

    printf(" Wattage of appliance: ");
    scanf("%f",&watts);

    printf(" Hours of use per day: ");
    scanf("%f",&hours);

    for(curr = dict; curr->next != NULL; curr=curr->next)
    {
        printf("comparing %s with %s\n", curr->name,state);  /* search for our state, set cents respectively */
        if((strcmp(curr->name,state)) == 0)
        {
            printf("State found! -- %s\n",state);
            cents=atof(curr->price);
            printf("%f cents\n",cents);
            break;
        } else {
            printf("state didn't match\n");
            continue;
        }
    }

    printf("\nAverage cost per day: %.2f\n",  calculate(watts,cents,hours));
    printf("Average cost per year: %.2f\n\n", (calculate(watts,cents,hours)) * 365);    

    return 0;

}

float calculate(float w, float c, float h)
{   
    float kwh = (w/1000) * h;
    float cos = kwh*(c/100);
    return(cos);
}

#包括
#包括
#包括
浮动计算（浮动w、浮动美分、浮动h）；
结构状态
{
字符*名称；
煤焦*价格；
结构状态*下一步；
};
int main（）
{
浮动瓦特，小时，美分；
int a=0；
结构状态*curr，dict[52]；
char st_ab[52][3]={“ID”、“ND”、“WA”、“AR”、“UT”、“LA”、“WY”、“SD”、“NE”、“KY”、“WV”、“OK”、“OR”、“MT”、“MS”、“TN”、“IN”、“NC”、“MS”、“IA”、“KS”、“VA”、“NM”、“MN”、“SC”、“AZ”、“GA”、“AL”、“CO”、“OH”、“TX”、“IL”、“FL”、“NV”、“MI”、“WI”、“PA”、“DC”、“DE”、“MA”、“RI”、“CA”、“ME”、“VT”、“NJ”、“NH”、“AK”、“CT”、“NY”、“HI”}；
[52][5][5][5][[5][[5][[5][5]{{{“8.0”、“8.0”、“8.0”、“8.8”、“8.9”、“8.9”、“8.0”、“8.0”、“8.0”、“8.0”、“8.0”、“8.0”、“8.1”、“8”、“8.0”、“8”、“8”、“8.1”、“8”、“8”、“8”、“8”、“8”、“8”、“8.7”、“8”、“8”、“8”、“8”、“8”、“8”、“8”、“8”、“8”、“8.9”、“8”、“8”、“8”、“8”、“8.9”、“8”、“8”、“8.9”、“8”、“8”、“8”、“8”、“8”、“8.9”、“8”、“8”、“8.9”、“8”、“8”、“8”、“8”、“8.9”、“8”、“8”、“8”、“8”、“8”、“8”、“8.9”、“8.9”、“8”、“8”、“8”、“8”、“8”、“8.9”、“8.9”、“8”、“8”、“8”、“8”、“8”、“8”、“8”、“第16.1条、第16.3条、第16.5条、第17.5条、第18.1条、，"18.1","33.2"};
半焦态[2]；
对于（a=0；anext！=NULL；curr=curr->next）/*用状态缩写和成本/KWh填写我们的列表*/
{
curr->name=st_ab[a]；
货币->价格=st_pr[a]；
a++；
}
puts（“已加载词典！”）；
printf（“两个字母的状态代码：”）；/*从用户处获取信息*/
scanf（“%s”，状态）；
printf（“设备瓦数：”）；
scanf（“%f”，瓦特）；
printf（“每天使用小时数：”）；
scanf（“%f”、&hours）；
for（curr=dict；curr->next！=NULL；curr=curr->next）
{
printf（“比较%s和%s\n”，curr->name，state）；/*搜索我们的状态，分别设置美分*/
如果（（strcmp（curr->name，state））==0）
{
printf（“找到状态！--%s\n”，状态）；
美分=atof（货币->价格）；
printf（“%f分\n”，分）；
打破
}否则{
printf（“状态不匹配\n”）；
继续；
}
}
printf（“\n每天平均成本：%.2f\n”，计算（瓦特、美分、小时）；
printf（“每年平均成本：%.2f\n\n”，（计算（瓦特、美分、小时））*365）；
返回0；
}
浮点数计算（浮点数w、浮点数c、浮点数h）
{   
浮充千瓦时=（w/1000）*小时；
浮动cos=千瓦时*（c/100）；
返回（cos）；
}

“curr->next从不为null”-那么，如果停止条件为curr->next等于null，您希望for循环如何停止？

问题在于

dict[a].next = &dict[a+1];

C标准保证

&dict[a+1]

永远不会为NULL，即使它超过了数组的末尾。这会导致下一个循环超过列表的末尾，并产生未定义的行为（程序可能会执行任何操作）。在第一个循环之后，执行

dict[51].next = NULL;

以结束链接列表

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编辑：使用

for（a=0；a
for（a=0；a）进行快速响应真是太棒了！经过其他一些小调整后，它现在工作得非常好！