我一直在为此程序苦苦挣扎。目的是在不使用libcsv的情况下读取简单的.csv文件。 经过研究和编程后,我想到了这个实现。它几乎就在那儿,但最终失败了。
我怀疑错误与str2uint64_t(ptr, &int64_converted, &error);
一致,但我不知道为什么。
为防万一,请根据我在此网页中找到的实现改编此实现: https://cboard.cprogramming.com/c-programming/47105-how-read-csv-file.html
顺便说一下,该程序可以编译为:
gcc -o q q.c && ./q file.csv
file.csv
可能类似于:
0,10,20,300,905
55,18,8,253,65
0,18,265,293,98
23,18,28,6675,86
677,20,28,293,100
实现:
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <errno.h>
void str2uint64_t(const char *str, uint64_t *intConverted, int *error)
{
// Converts string to uint64_t
*intConverted = 0;
*error = 0;
const char *s = str;
int sign = *s;
char *end;
errno = 0;
const uint64_t sl = strtoull(str, &end, 10);
if (end == str)
{
//fprintf(stderr, "%s: not a decimal number\n", str);
*error = 1;
}
else if ('\0' != *end)
{
//fprintf(stderr, "%s: extra characters at end of input: %s\n", str, end);
*error = 1;
}
else if (ERANGE == errno)
{
//fprintf(stderr, "%s out of range of type uint64_t\n", str);
*error = 1;
}
else if (sign == '-')
{
//fprintf(stderr, "%s negative\n", 0);
//errno = ERANGE;
*error = 1;
}
//return sl;
*intConverted = sl;
}
void *newMatrix(size_t rows, size_t cols)
{
return malloc (sizeof(uint64_t[rows][cols]));
}
void importMatrix(char CSVFilePath[], size_t rows, size_t cols, uint64_t matrix[rows][cols])
{
size_t i, j;
uint64_t int64_converted;
int error = 0;
FILE *CSVfile = fopen(CSVFilePath, "r");
if (CSVfile == NULL)
{
perror("Error");
exit(EXIT_FAILURE);
}
char buffer[BUFSIZ], *ptr;
for (i = 0; fgets(buffer, sizeof buffer, CSVfile); ++i)
{
for (j = 0, ptr = buffer; j < rows; ++j, ++ptr)
{
str2uint64_t(ptr, &int64_converted, &error);
if (error == 0)
{
// From https://cboard.cprogramming.com/c-programming/47105-how-read-csv-file.html >> array[i][j] = (int)strtol(ptr, &ptr, 10);
matrix[i][j] = int64_converted;
}
else
{
printf("Failed to import matrix\n");
exit(0);
}
}
}
fclose(CSVfile);
putchar('\n');
}
int main(int argc, char *argv[])
{
if (argc < 2)
{
fprintf(stderr, "Usage: ./<program> <file.csv>\n");
exit(EXIT_FAILURE);
}
size_t rows = 5;
size_t cols = rows;
uint64_t (*matrix)[rows] = newMatrix(rows, cols);
importMatrix(argv[1], rows, cols, matrix[rows][cols]);
//////////////////////////////
return 0;
}
答案 0 :(得分:1)
通过geany使用gcc发布的代码会导致:
gcc -ggdb -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled.c"
untitled.c: In function ‘main’:
untitled.c:114:39: warning: passing argument 4 of ‘importMatrix’ makes pointer from integer without a cast [-Wint-conversion]
importMatrix(argv[1], rows, cols, matrix[rows][cols]);
^~~~~~
untitled.c:58:6: note: expected ‘uint64_t (*)[(sizetype)(cols)]’ but argument is of type ‘uint64_t {aka long unsigned int}’
void importMatrix(char CSVFilePath[], size_t rows, size_t cols, uint64_t matrix[rows][cols])
^~~~~~~~~~~~
Compilation finished successfully
这两个警告都是严重的。结果是编译器生成了错误的代码。请修正这些警告并编辑您的问题。
此语句的功能是:main()
:
importMatrix(argv[1], rows, cols, matrix[rows][cols]);
将第四个参数的内存内容传递给数组末尾1:matrix[][]
该语句应为:
importMatrix( argv[1], rows, cols, matrix );
OT:关于此声明:
fprintf(stderr, "Usage: ./<program> <file.csv>\n");
该程序可以任何名称运行。因此建议声明为:
fprintf(stderr, "Usage: %s <file.csv>\n", argv[0]);
然后将显示实际的执行名称
OT:关于此声明:
return malloc (sizeof(uint64_t[rows][cols]));
强烈建议使用:
return malloc( sizeof( uint64_t ) * rows * cols );
然后在调用importMatrix()
后进入函数newMatrix()
此声明:
uint64_t (*matrix)[rows] = newMatrix(rows, cols);
不正确,因为对newMatrix()
的调用未返回指针数组,因此该语句应类似于:
typedef matrixType uint64_t matrix[rows][cols];
matrixType * matrix = newMatrix(rows, cols);