分配给变量[]时“在“]”标记之前解析错误答案

【问题标题】："Parse error before "]" token" when assigning to variable[]分配给变量[]时“在“]”标记之前解析错误
【发布时间】：2015-04-21 17:13:03
【问题描述】：

我的 C 程序如下。它试图从文件中读取一组值，并创建一个表格来生成一些浮点数的平均值。

int main(void)
{

    int lake, beach, samples, count, sum;
    float e_coli, ave;
    char recc[20], lakename[20], beachname[20];
    FILE *in;
    sum = 0;
    count = 0;
    in = fopen("july15.data", "r");
    printf("Lake   |   Beach   | Avg E Coli Level |  Reccomendation |\n");
    while (1)
    {
        fscanf(in, "%d, %d, %d", &lake, &beach, &samples);
        count = samples;
        while (count < samples)
        {
            fscanf(in, "%f", e_coli);
            sum = sum + samples;
        }
        ave = sum / samples;
        if (lake == 1)
            lakename[] = "Ontario";
        else if (lake == 2)
            lakename[] = "Erie";
        else if (lake == 3)
            lakename[] = "Huron";
        else if (lake == 4)
            lakename[] = "Muskoka";
        else if (lake == 5)
            lakename[] = "Simcoe";
        if (beach == 100)
            beachname[] = "Kew Beach";
        else if (beach == 101)
            beachname[] = "Sunnyside Beach";
        else if (beach == 102)
            beachname[] = "Sandbanks";
        else if (beach == 201)
            beachname[] = "Port Dover";
        else if (beach == 202)
            beachname[] = "Port Burwell";
        else if (beach == 203)
            beachname[] = "Crystal Beach";
        else if (beach == 301)
            beachname[] = "Goderich";
        else if (beach == 302)
            beachname[] = "Sauble Beach";
        else if (beach == 303)
            beachname[] = "Kincardine";
        else if (beach == 401)
            beachname[] = "Muskoka Beach";
        else if (beach == 501)
            beachname[] = "Sibbald Point";
        if (samples < 3)
            recc[] = "INSUFFICENT DATA";
        else if (samples > 3 && ave < 50)
            recc[] = "OPEN";
        else if (samples > 3 && ave > 50)
            recc[] = "CLOSED";
        printf("%s,      %s,      %f,      %s\n", lakename, beachname, ave, recc);
    }
    return(0);
}

不幸的是，它在第 22、32 和 54 行产生了错误 parse error before "]" token。实际上，这些是定义字符串的第一个区域。所以第22行是lakename[] = "Ontario"第32行是beachname[] = "Kew Beach"，第54行是recc[] = "INSUFFICIENT DATA"。

我尝试将空格数插入[]（即lakename[9] = "Ontario"），但随后出现错误"assignment makes integer from pointer without cast"。

【问题讨论】：

要复制/连接 C-“字符串”，请使用 strcpy() (man7.org/linux/man-pages/man3/strncpy.3.html) 或 strcat(): (man7.org/linux/man-pages/man3/strcat.3.html)。
另外请正确缩进您的代码。

标签： c arrays string loops while-loop

【解决方案1】：

数组在 C 中是不可赋值的。

有效的赋值/初始化是

char lakename[] = "Ontario";

你得到错误

lakename[9] = "Ontario"

因为lakename[9] 只能容纳一个字符。

无论如何，还有其他选项可以复制 strcpy() 和家庭等字符串

【讨论】：

char lakename[] = "Ontario"; 被恰当地称为“initialisation”。

【解决方案2】：

lakename[] = "Ontario"; 或 beachname[] = "Kew Beach"; 形式的语句是无效的 C。为了安全起见，请改用 strcncpy(beachname, "Kew Beach", sizeof(beachname)-1); 之类的东西，在上面的某个地方使用 beachname[sizeof(beachname)-1] = 0;。或者，将变量设为char *，然后将它们分配为指针。在这种情况下，这更合适。

【讨论】：

【解决方案3】：

假设您的输入数据看起来像下面的示例，该解决方案应该可以工作。

1 101 5 5.2 10.9 20.1 30.56 80.6
1 301 4 50.0 30.3 40.2 20.4

代码解决方案。

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

int main (void){
    int lake, beach, samples;
    char recc[20], lakename[20], beachname[20];

    FILE *in;
    in = fopen ("july15.data", "r");
    if(in == NULL) {
        exit(EXIT_FAILURE); 
    }

    char *line = NULL;
    size_t len = 0;
    ssize_t read;
    
    printf ("Lake   |   Beach   | Avg E Coli Level |  Reccomendation |\n");

    while ((read = getline(&line, &len, in)) != -1) {
        // Count number of spaces
        double sum = 0;
        int tokens = 0;
        int i = 0;
        for (; i < len; i++) {
            if (line[i] == ' ') {
                tokens++;
            }
        }
        // Add one extra token
        tokens = tokens + 1;
        if (tokens < 4) {
            continue;
        }

        i = 0;
        // Convert tokens to floats
        double *values = malloc(sizeof(double) * tokens);
        char *token = strtok(line, " ");  
        while (token != NULL) {
            values[i++] = atof(token);
            token = strtok(NULL, " ");
        }

        lake = values[0];
        beach = values[1];
        samples = values[2];
        for (i = 3; i < tokens; i++) {
            sum = sum + values[i];
        }
        free(values);

        double ave = sum / samples;
        
        if (lake == 1)
            strcpy(lakename, "Ontario");
        else if (lake == 2)
            strcpy(lakename, "Erie");
        else if (lake == 3)
            strcpy(lakename, "Huron");
        else if (lake == 4)
            strcpy(lakename, "Muskoka");
        else if (lake == 5)
            strcpy(lakename, "Simcoe");

        if (beach == 100)
            strcpy(beachname, "Kew Beach");
        else if (beach == 101)
            strcpy(beachname, "Sunnyside Beach");
        else if (beach == 102)
            strcpy(beachname, "Sandbanks");
        else if (beach == 201)
            strcpy(beachname, "Port Dover");
        else if (beach == 202)
            strcpy(beachname, "Port Burwell");
        else if (beach == 203)
            strcpy(beachname, "Crystal Beach");
        else if (beach == 301)
            strcpy(beachname, "Goderich");
        else if (beach == 302)
            strcpy(beachname, "Sauble Beach");
        else if (beach == 303)
            strcpy(beachname, "Kincardine");
        else if (beach == 401)
            strcpy(beachname, "Muskoka Beach");
        else if (beach == 501)
            strcpy(beachname, "Sibbald Point");
        
        if (samples < 3)
            strcpy(recc, "INSUFFICENT DATA");
        else if (samples > 3 && ave < 50)
            strcpy(recc, "OPEN");
        else if (samples > 3 && ave > 50)
            strcpy(recc, "CLOSED");

        printf("%s,      %s,      %f,      %s\n", lakename, beachname, ave, recc);
    }

    free(line);
    fclose(in);

    return(0);
}

【讨论】：

sizeof 的用法没有任何意义。
这仍然没有意义，因为strncpy() 的第三个参数应该用于保护，而不是溢出目标缓冲区，就像这样（假设dst 是char 和不仅仅是一个指针）memset(dst, 0, sizeof dst); strncpy(dst, src, sizeof dst -1);供您参考：man7.org/linux/man-pages/man3/strncpy.3.html
嘿，你说得对。我应该更加注意这一点。我将 strncpy 更改为 strcpy 并假设缓冲区始终可以适合复制的字符串。再次感谢！
好的，谢谢大家。但输入数据看起来不像那样。相反，它在一行中。这是第一行：4 401 4 30.3 50.4 50.5 60.6 其中 4 是湖，401 是海滩，4 是样本数，其余 4 个浮点数是生态读数...
它看起来怎么样？

【解决方案4】：

我会进行一些修改，使代码更具可读性，并提醒用户如果不支持没有湖泊或没有海滩

#include <stdio.h>

int main (void){
    int lake, beach, samples, count, sum;
    float e_coli, ave;
    char *recc;
    char *lakename;
    char *beachname;
    FILE *in;

    sum = 0;
    count = 0;
    in = fopen ("july15.data", "r");
    printf ("Lake   |   Beach   | Avg E Coli Level |  Reccomendation |\n");

    int iterator;
    char lake_names[6][10]={
        {"Ontario"},
        {"Erie"},
        {"Huron"},
        {"Muskoka"},
        {"Simcoe"},
        {"No lake"},
    };

    int beach_code[12] ={100,101,102,201,202,203,301,302,303,401,501,0};

    char beach_names[12][20]={
        {"Kew Beach"},
        {"Sunnyside Beach"},
        {"Sandbanks"},
        {"Port Dover"},
        {"Port Burwell"},
        {"Crystal Beach"},
        {"Goderich"},
        {"Sauble Beach"},
        {"ncardine"},
        {"Muskoka Beach"},
        {"Sibbald Point"},
        {"No beach"},
    };

    char reccs[3][20]={
        {"INSUFFICENT DATA"},
        {"OPEN"},
        {"CLOSED"},
    };




    while (fscanf (in, "%d, %d, %d", &lake, &beach, &samples) !=  EOF)
    {
        recc=&reccs[0][0];
        lakename=&lake_names[5][0];
        beachname=&beach_names[11][0];

        count = samples;

        while (count < samples)
        {
            fscanf (in, "%f",&e_coli);
            sum = sum + samples;
        }

        ave = sum / samples;

        for(iterator=1;iterator<5;iterator++)
        {
            if(lake==iterator)
                lakename=&lake_names[iterator][0];
        }

        for(iterator=0;iterator<11;iterator++)
        {
            if(beach==beach_code[iterator])
                beachname=&beach_names[iterator][0];
        }

        if (samples < 3)
            recc = &reccs[0][0];
        else if (samples > 3 && ave < 50)
            recc = &reccs[1][0];
        else if (samples > 3 && ave > 50)
            recc = &reccs[2][0];
        printf("%s,      %s,      %f,      %s\n", lakename, beachname, ave, recc);
    }
    return(0);
}

【讨论】：