用C语言从文件中读取两个不同的列并将它们存储在两个不同的数组中？

我不太熟悉C中的文件处理。我有一个.txt文件，其中包含以下内容：

135.33208225150605 OK
165.1233490789463 OK
245.2329542568199 OK
301.9041144959863 D

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
for(i=0;i<=148;i++){

    fgets(buff,sizeof(buff),fp);
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    line[i] = buffcopy;
    }

   fclose(fp);

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
double dv;

for(i=0;i<=148 && fscanf(fp, "%lf %[^\n]", &dv, buff)==2;i++){
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    string_array[i] = buffcopy;//char *string_array[149];
    numbers[i] = dv;           //double numbers[149];
}
fclose(fp);

我想将带有数字的列存储在一个双数组中，将字符串列存储在另一个字符串数组中

我是这样做的：

135.33208225150605 OK
165.1233490789463 OK
245.2329542568199 OK
301.9041144959863 D

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
for(i=0;i<=148;i++){

    fgets(buff,sizeof(buff),fp);
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    line[i] = buffcopy;
    }

   fclose(fp);

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
double dv;

for(i=0;i<=148 && fscanf(fp, "%lf %[^\n]", &dv, buff)==2;i++){
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    string_array[i] = buffcopy;//char *string_array[149];
    numbers[i] = dv;           //double numbers[149];
}
fclose(fp);

文件*fp；
fp=fopen（“协议1QT.序列”，“r”）；
int i；
对于（i=0；i可以用
strtok（）
将其分开，或者如果它只是这个简单的例子，可以像这样使用
sscanf（）

char text[10]; // more if it could be larger
float number;
if (sscanf(line, "%f%9s", &number, text) == 2)
    process_columns(number, text);

另外：在
fgets（）
之前，检查
fp
中的
NULL
，并将
fgets（）
置于
条件中。
类似如下：

135.33208225150605 OK
165.1233490789463 OK
245.2329542568199 OK
301.9041144959863 D

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
for(i=0;i<=148;i++){

    fgets(buff,sizeof(buff),fp);
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    line[i] = buffcopy;
    }

   fclose(fp);

FILE *fp;
fp = fopen("protocol1QT.seq", "r");
int i;
double dv;

for(i=0;i<=148 && fscanf(fp, "%lf %[^\n]", &dv, buff)==2;i++){
    char *buffcopy = malloc(strlen(buff) + 1);
    if(buffcopy == NULL) {fprintf(stderr, "out of memory\n"); exit(1); }
    strcpy(buffcopy, buff);
    string_array[i] = buffcopy;//char *string_array[149];
    numbers[i] = dv;           //double numbers[149];
}
fclose(fp);

文件*fp；
fp=fopen（“协议1QT.序列”，“r”）；
int i；
双dv；
对于（i=0；i