使用printf我使用的函数的输出返回错误的值

时间:2021-01-13 19:55:45

I have this code:

我有这个代码:

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

char *K2G(int k)
{
    static char g[10];
    if (k > 1048576) {
        sprintf(g, "%.2fGB", (float) k / 1048576);
    } else {
        if (k > 1024) {
            sprintf(g, "%.2fMB", (float) k / 1024);
        } else {
            sprintf(g, "%dKB", k);
        }
    }
    printf("%s\n", g);
    return g;
}

main()
{
    FILE *fp;
    int imt = 0, imf = 0, imu = 0;
    char cmt[40], cmf[40], cmti[20], cmfi[20], a[20], b[20];
    while (1) {
        system("clear");
        fp = fopen("/proc/meminfo", "r");
        fgets(cmt, 40, fp);
        fgets(cmf, 40, fp);
        fclose(fp);
        printf("%s%s\n", cmf, cmt);
        sscanf(cmt, "%s%d%s", a, &imt, b);
        sscanf(cmf, "%s%d%s", a, &imf, b);
        imu = imt - imf;
        printf("%s/%s=%d%\n", K2G(imu), K2G(imt), imu * 100 / imt);
        sleep(1);
    }
}

The output I get is something like:

我得到的输出是这样的:

MemFree:          494256 kB
MemTotal:       10258000 kB

9.78GB
9.31GB
9.31GB/9.31GB=95%

The last line always displays the same two values before the equal sign. The output should have been:

最后一行始终在等号前显示相同的两个值。输出应该是:

MemFree:          494724 kB
MemTotal:       10258000 kB

9.31GB
9.78GB
9.31GB/9.78GB=95%

Why do I get duplicate values when I call function K2G with printf? This is the line giving me the incorrect results:

当我用printf调用函数K2G时,为什么会出现重复值?这条线给我的结果不正确:

printf("%s/%s=%d%\n", K2G(imu), K2G(imt), imu * 100 / imt);

1 个解决方案

#1


2  

This behaviour is normal, you return a pointer to g which is a static buffer, and on each call this buffer will be overwritten.

这种行为是正常的,你返回一个指向g的指针,这是一个静态缓冲区,并且在每次调用时,这个缓冲区都会被覆盖。

So if you do printf(..., K2G(x), K2G(Y),...), the parameters "seen" by printf will both be the same g buffer with it's latest content.

因此,如果你做printf(...,K2G(x),K2G(Y),...),printf“看到”的参数将与它的最新内容是相同的g缓冲区。

You can do this:

你可以这样做:

char simu[20];
char simt[20];
strcpy(simu, K2G(imu));
strcpy(simt, K2G(imt));
printf(..., simu, simt,...);

EDIT:

编辑:

Or you can use another pattern where you have to provide a buffer to K2G:

或者您可以使用另一种模式,您必须为K2G提供缓冲区:

char *K2G(int k, char *g)
{
    if (k > 1048576) {
        sprintf(g, "%.2fGB", (float) k / 1048576);
    } else {
        if (k > 1024) {
            sprintf(g, "%.2fMB", (float) k / 1024);
        } else {
            sprintf(g, "%dKB", k);
        }
    }
    printf("%s\n", g);
    return g;
}

...

char simu[20];
char simt[20];
K2G(imu, simu);
K2G(imt, simt);
printf(..., simu, simt,...);

This is more transparent and avoids the usage of strcpy.

这更透明,避免使用strcpy。

#1


2  

This behaviour is normal, you return a pointer to g which is a static buffer, and on each call this buffer will be overwritten.

这种行为是正常的,你返回一个指向g的指针,这是一个静态缓冲区,并且在每次调用时,这个缓冲区都会被覆盖。

So if you do printf(..., K2G(x), K2G(Y),...), the parameters "seen" by printf will both be the same g buffer with it's latest content.

因此,如果你做printf(...,K2G(x),K2G(Y),...),printf“看到”的参数将与它的最新内容是相同的g缓冲区。

You can do this:

你可以这样做:

char simu[20];
char simt[20];
strcpy(simu, K2G(imu));
strcpy(simt, K2G(imt));
printf(..., simu, simt,...);

EDIT:

编辑:

Or you can use another pattern where you have to provide a buffer to K2G:

或者您可以使用另一种模式,您必须为K2G提供缓冲区:

char *K2G(int k, char *g)
{
    if (k > 1048576) {
        sprintf(g, "%.2fGB", (float) k / 1048576);
    } else {
        if (k > 1024) {
            sprintf(g, "%.2fMB", (float) k / 1024);
        } else {
            sprintf(g, "%dKB", k);
        }
    }
    printf("%s\n", g);
    return g;
}

...

char simu[20];
char simt[20];
K2G(imu, simu);
K2G(imt, simt);
printf(..., simu, simt,...);

This is more transparent and avoids the usage of strcpy.

这更透明,避免使用strcpy。